Representation theory

Fredrik Meyer

Abstract

These are notes from the course MAT4270 on representation the-ory, autumn 2015. The lectures were held by Sergey Neshyevey. Thenotes are mine, as well as any mistakes. The first half is about rep-resentations of finite groups, the second half about representations ofcompact Lie groups.

1 Introduction to representation theoryIn the first part, G is always a finite group and V a finite-dimensionalvector space over the complex numbers. Most results will hold overany algebraically closed field of characteristic zero.

1.1 Motivating exampleIn 1896 Dedekind made the following observation. Let C[xg | g ∈ G]be the free C-algebra with basis the elements of G. Then one can formthe following polynomial:

PG(xρ1 , . . . , xρn) = det((xρigj )

ni,j=1

).

Note that the matrix is just the multiplication table for the groupG if we identify the variable xg with the group element g. The problemis to decompose PG into irreducible polynomials.

Example 1.1. Let G = S3. Then G is generated by two elements rand s with the relation srs = r2. If we write the elements of G as

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e, r, r2, s, sr, sr2, then the multiplication table looks like

e r r2 s sr sr2

r r2 e sr2 s sr

r2 e r sr sr2 s

s sr sr2 e r r2

sr sr2 s r2 e r

sr2 s sr r r2 e

.

The determinant is quite long, so I won’t write it out. Suffice it to sayit is a degree 6 polynomial with 146 terms. Using a computer algebrasystem such as Macaulay2, one can decompose it. It decomposes intothree factors: one quadratic and two linear. They are the following(we change notation to avoid confusion about exponents):

xe + xr + xr2 + xs + xsr + xsr2

andxe + xr + xr2 − xs − xsr − xsr2

and

x2e−xexr+xr2−xexr2−xrxr2+x2

r2−x2s+xsxsr−x2

sr2+xsxsr2−x2sr2−x2

sr2 .

We will see later that the first factor corresponds to the trivial repre-sentation, the second to the alternating representation, and the thirdis the so-called standard representation of G. F

Frobenius proved the following theorem:

Theorem 1.2. Let PG =∏ri=1 P

mii be the decomposition of PG into

irreducibles. Then

1. mi = degPi for every i.

2. r is the number of conjugacy classes in G.

In particular, since r = |G| if and only if G is abelian, PG decomposesinto linear factors if and only if G is abelian.

In trying to prove the above theorem, Frobenius basically had todevelop representation theory.

1.2 RepresentationsA representation of G on V is a homomorphism π : G → GL(V ).We will denote a representation interchangably by π, (π, V ), Vπ or V ,

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depending upon the situation. We sometimes also say that V is aG-module.

Thus a representation is a linear action of G on V , so that theaction of G on V always can be represented as some group of matrices.This matrix group is isomorphic to G if the action is faithful.

Example 1.3. The trivial representation ε of any group G is givenby letting V = C and ε(g) = e for any g ∈ G. F

Example 1.4. We define the left regular representation. Let C[G]be the space of functions on G. It is a finite-dimensional vector spaceunder pointwise addition. Define λ : G→ GL(C[G]) by

g 7→(f 7→ (h 7→ f(g−1h))

)This vector space have a basis consisting of the characteristic functionseg for g ∈ G. On these elements on sees that the action is given byg · eh = egh.

Thus any group have at least one non-trivial representation. F

Example 1.5. Similarly, one has the permutation representation.Let X be a G-set, that is, a set on which G act by permutations. Thenone forms the space C[X] of C-valued functions on X. It has a basisex of characteristic functions, and the action of G on C[X] is given bygex = egx. F

As in any reasonable category, there is a notion of isomorphism ofrepresentations. An equivalence of representations (π, V ) and (θ,W )is given by a vector space isomorphism T : V →W such that T (g ·v) =g · T (v) for all g ∈ G.

A map T : V → W satisfying the last condition above is calledan intertwiner of V and W . The set of intertwiners is denoted byMor(π, θ) or HomG(V,W ).

A subspace W ⊂ V is invariant if g · W ⊂ W for all g ∈ G.Letting θ(g) = π(g)

∣∣W, we get another representation of G, called a

subrepresentation of G. We write π∣∣W

for θ.If we have two representation (π, V ) and (π′, V ′), we can form the

direct sum representation by letting g ∈ G act on V ⊕ V ′ compo-nentwise. Note that π is a subrepresentation of V ⊕ V ′.

The following proposition has important consequences. Note thatthe proof works for any field of characteristic not dividing |G|.

Proposition 1.6 (Mascke’s theorem). Let (π, V ) be a representationof G and W ⊂ V an invariant subspace. Then there exists a com-plementary invariant subspace W⊥. That is, W is also invariant andsuch that V = W ⊕W⊥.

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Proof. We prove this by “averaging”, a process we will do again andagain. Let P : V →W be any projection from V to W . Then define

Π(v) =1

|G|∑g∈G

gP (g−1v).

This is a G-linear morphism, because

Π(h · v) =1

|G|∑g∈G

gP (g−1h · v)

=1

|G|∑g∈G

gP ((h−1g) · v)

=h

|G|∑g∈G

h−1gP ((h−1g) · v) = h ·Π(v).

It is also surjective, since if v ∈W , then clearly Π(v) = v. Hence Π is aG-linear surjection ontoW . Then it is clear that ker Π is a G-invariantsubspace of V complementary to W .

We say that a representation is irreducible if it has no properinvariant subspaces. A representation is completely reducible if itdecomposes into a direct sum of irreducible representations.

Example 1.7. Let G = (R,+) act on R2 by the matrices1 a

0 1

.

This leaves the x-axis fixed, so the representation is not irreducible.But it does not split into irreducible representations, since there is nocomplementary subspaces. F

Example 1.8. The same example with G = (Fp,+) and V = F2p

works. Again, the space spanned by the vector (1, 0)T is invariant, soV is not irreducible. The same computation as in the previous exampleshows that are no complementary invariant subspace. F

Example 1.9. Let G = Z/2 act on V = F2p by (x, y) 7→ (y, x). Then

the diagonal ∆ is an invariant subspace. Letting P : F2p → F2

p be themap (x, y) 7→ (x, x) be a projection onto ∆, going through the proof,we see that

Π((x, y)) =1

2((x, x) + (y, y)) =

1

2(x+ y, x+ y)

is an intertwiner. The kernel is spanned by (1,−1)T if p 6= 2, so we seethat V is completely reducible if and only if p 6= 2. F

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However, we don’t worry, because we work over C. Here things arenicer:

Theorem 1.10. Any finitely-dimensional representation of a finitegroup G is completely reducible.

Proof. The proof is a direct consequence of Maschke’s theorem (??).Just inductively find invariant subspaces. Since the representation isfinite-dimensional, the process must stop.

We next prove Schur’s lemma, which although it has a very simpleproof, has important consequences.

Proposition 1.11 (Schur’s lemma). Let (V, π) and (W, θ) be irre-ducible representations. Then

1. If π 6∼ θ, then HomG(V,W ) = 0.

2. If π ∼ θ, then HomG(V,W ) is 1-dimensional.

Proof. Both the kernel and the image of a morphism of representationsis a representation. But since V and W are irreducible, it follows thatkerϕ = 0 or kerϕ = V for any ϕ : V →W . This proves i).

Now suppose Π : V →W is an equivalence of representations. Letϕ ∈ HomG(V,W ) be non-zero. Then ϕ must be an equivalence aswell, since the neither the kernel nor image can be non-zero. ConsiderT = ϕ−1 Π : V → V . This is a linear map, and since we work overC, it has an eigenvalue λ. Now consider T −λ · idV . This has non-zeroinvariant kernel, hence T − λidV = 0.

Proposition 1.12. If G is abelian, then any irreducible representationof G is 1-dimensional.

Proof. Each g ∈ G gives a map π(g) : V → V ∈ End(V ). This is anintertwining map, because π(g)π(h) = π(gh) = π(hg) = π(h)π(g).

If π is not trivial, by Schur’s lemma, this map is just multiplicationby some constant. But any 1-dimensional subspace of V is invariantunder this action, so V must be 1-dimensional in order to be irre-ducible.

Thus, up to equivalence, every irreducible representation of a finiteabelian group is just a homomorphism χ : G → C. But since G isfinite, this actually maps into T := z ∈ C | |z| = 1. Such maps arecalled characters of G.

Denote by G the set of irreducible representations of G. We callG the Pontryagin dual of G. It is an abelian group under pointwisemultiplication.

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Remark. If G is non-abelian, there will always exist irreducible rep-resentations of dimension ≥ 2. This follows because the regular repre-sentation G→ GL (C[G]) is injective.

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2 Density and orthogonality

2.1 Density theoremsLet (V, π) be an irreducible representation. The set of operators π(g)for g ∈ G is quite large. By definition, if x 6= 0, then

V = Spanπ(g)x | g ∈ G.

By Schur’s lemma, we also know that

End(π) = T ∈ End(V ) | Tπ(g) = π(g)T ∀ g ∈ G = C · idV .

Remark. Over any field, End(π) is a division algebra over that field.But if the field is algebraically closed, any division algebra is isomorphicto the field itself.

Theorem 2.1 (The density theorem). Assume π1, . . . , πn are pairwiseinequivalent irreducible representations, with Vi = Vπi . Consider thedirect sum representation V1 ⊕ . . .⊕ Vn.

ThenSpanπ(G) = End(V1)⊕ . . .⊕ End(Vn).

Proof. Read the proof of Theorem ?? first. First break (V, π) intoirreducible representations Vi. It follows directly from Schur’s lemmathat π(G)′ = CidV1

⊕ . . .⊕ CidVn .Now, the elements of CidV1

⊕ . . .⊕CidV2are block diagonal scalar

matrices. It is an easy exercise with matrices to see that the commu-tator of this set is just

End(V1)⊕ . . .⊕ End(Vn).

This proves the theorem.

First we need some notation. Let V be some vector space andX ⊂ End(V ) a subset of the endomorphisms of V . Let

X ′∆= T ∈ End(V ) | TS = ST for all S ∈ X.

We call X ′ the commutant of X. Thus if (V, π) is a representation,then by definition:

End(π) = π(G)′.

Theorem 2.2. For any finite-dimensional representation (V, π) wehave

π(G)′′ = Spanπ(G).

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Proof. The inclusion ⊇ is clear, since π(G)′′ = End(π)′, and we haveπ(g)S(v) = π(g)(Sv) = S(π(g)v) for all S ∈ End(π).

For the other inclusion, let T ∈ π(G)′′. Let x1, . . . , xn be a basisof V . Consider the representation θ = π ⊕ . . . ⊕ π (n times) on W =⊕ni=1V . Let S be the operator T ⊕ . . .⊕ T on W .

Then S ∈ θ(G)′′, because

(Sθ)(v) = S(θ · (v1, . . . , vn))

= S(π(v1), . . . , π(vn))

= (Tπ(v1), . . . , Tπ(vn))

= (π(Tv1), . . . , π(Tvn))

= θ · (Tv1, . . . , T vn)

= θSv.

Let x = (x1, . . . , xn) ∈ W . Then Span θ(G)x is an invariant sub-space of W . Then there exists an invariant complementary subspace.Let P : W → Span θ(G)x be the projection with kernel that invariantsubspace, so that P ∈ Θ(G)′ (that is, P is a G-morphism). Then

PSx = SPx = Sx

since Px ∈ Span θ(G)x. Hence Sx ∈ Span θ(G)x as well. This meansthat there exist αg ∈ C such that

Sx =∑g

αgθ(g)x.

This is the same as saying that

Txi =∑g

αgπ(g)xi,

hence T =∑αgπ(g), so T ∈ Spanπ(G).

Remark. Note that any irreducible representation is contained in theregular representation. Indeed, let (V, π) be irreducible, and pick somenon-zero x ∈ V . Then the map

C[G]·x−→ V

must be surjective, and by for example Schur’s lemma it has a section.Thus the density theorem says that

|G|2 =∑π∈G

|Vi|2.

In particular, there are only finitely many isomorphism classes of rep-resentations of G.

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2.2 Orthogonality relationsLet (V, π) be a representation and let ρ ∈ V ∗ = HomC(V,C). Thefunction απρ,x on G defined by απρ,x(g) = ρ(π(g)x) is called a matrixcoefficient of π. If we fix a basis e1, . . . , en of V and consider the dualbasis e∗1, . . . , e∗n of V ∗, then we write απij instead of απe∗i ,ej .

In the basis e1, . . . , en we have

π(g) =(απij(g)

)ij∈ GL(V ).

The following theorem is usually called the orthogonality rela-tions:

Theorem 2.3. We have:

1. If π and θ are inequivalent irreducible representations, then

1

|G|∑g∈G

απf,x(g)αθρ,y(g−1) = 0

for all f ∈ V ∗π , ρ ∈ V ∗θ , x ∈ Vπ and y ∈ Vθ.2. If π is irreducible, then

1

|G|∑g

απf,x(g)απρ,y(g−1) =f(y)ρ(x)

dimπ.

Proof. It is the averaging trick again. First observe that if T : Vθ → Vπis any linear operator, then

S =1

|G|∑g∈G

π(g)Tθ(g−1)

is in HomG(θ, π) by averaging. Thus, since π and θ are inequivalent,it follows from Schur’s lemma that this must be zero.

Now let T (v) = ρ(v)x. This is a linear operator. Then

0 = f(Sy) = f

1

|G|∑g∈G

π(g)Tθ(g−1)(y)

=

1

|G|∑g∈G

f(π(g)ρ(θ(g−1)y)x)

=1

|G|∑g∈G

f(π(g)x)ρ(θ(g−1)y)

=1

|G|∑g∈G

απf,x(g)αθρ,y(g−1)

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This proves part 1.Now suppose π = θ. Then S ∈ End(π) = C · idV . Thus S = α · idV

for some α ∈ C. Taking traces on both sides, we get α = TrT/ dimπ(note that T and S have the same trace).

Then

1

|G|∑g

απf,x(g)απρ,y(g−1) = f(S(y)) = f(y)TrT

dimπ=f(y)ρ(x)

dimπ,

since TrT = ρ(x).

Let C[G] denote the vector space of functions G → C. Instead ofwriting expressions like

∑g αgπ(g) it is convenient to introduce the

following notation. Define the convolution of two functions f1, f2 ∈C[G] by

(f1 ∗ f2)(g) =∑

g=g1g2

f1(g1)f2(g2) =∑h∈G

f1(h)f2(h−1g).

We have δg1 ∗ δg2 = δg1g2 (easy check!). The convolution product isassociative and makes C[G] into an algebra, called the group algebraof G.

Note that with this product, C[G] could just as well have beendefined as the associative algebra C〈xg | g ∈ G〉 module the relationsdefined by xgxh = xgh.

Given a representation π : G → GL(V ) we can define an algebrahomomorphism C[G] → End(V ) by π(f) =

∑g∈G f(g)π(g). Thus

V becomes a left C[G]-module. Conversely, given any unital algebrahomomorphism π : C[G] → End(V ), we get a representation by g 7→π(δg). Thus we have a correspondence between representations of Gand left C[G]-modules. Using these notions, the Density Theorem(Theorem ??) becomes π(C[G]) = ⊕i End(Vi).

Formalizing Remark ??, we get:

Theorem 2.4. Let G be the set of equivalence classes of irreduciblerepresentations of G. Then G is finite and

∑[π]∈G(dimπ)2 = |G|.

Furthermore, we have

λ ∼⊕

[π]∈G

π⊕ dimπ.

Here λ is the left regular representation of G.

Proof. The first two statements are already proved in the Remark.The only thing to be proved is the part about the multiplicity of π inλ.

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Note that as a representation, C[G] is just the left regular repre-sentation of G. Note also that EndVi ' V dimV : let v1, . . . , vn bea basis of V . Then the map End(Vi) 3 ϕ 7→ (ϕ(v1), . . . , ϕ(vn)) is anisomorphism of representations.

Then the density theorem implies the statement.

Example 2.5. Let G = S3, the symmetric group on 3 elements. Wehave that |G| = 6. We have two one-dimensional representations: thetrivial representation and the sign representation g 7→ sign(g) ∈ ±1.Hence 6 = 12 + 12 +n2

1 +n22 + . . ., where ni is the dimension of the ith

irreducible representation. This is only possible if n1 = 2 and ni = 0for i > 1. Thus there are only three irreducible representations of S3.

Let G = 〈ρ, σ〉 with ρ3 = e, σ2 = e and σρσ = ρ−1. The uniqueirreducible 2-dimensional representation of S2 can be realized as theaction on C2 given by

ρ 7→

sin(

2π3

)− sin

(2π3

)cos(

2π3

)cos(

2π3

)

and

σ 7→

−1 0

0 1

.

The density theorem here says that

C[G] = eC⊕ . . .

[[sett inn dekomposisjon av C[G] som algebra]] F

We will now motivate the name "orthogonality relations". First ofall, note that if Vπ is a representation of a finite group G, and 〈, 〉′ isany Hermitian scalar product on Vπ, then

〈v, w〉 :=1

〈G〉∑g∈G〈π(g)v, π(g)w〉′

is another Hermitian scalar product on Vπ, making the representationunitary, meaning that the inner product is invariant under the actionof the group.

If π is any unitary representation, choose an orthonormal basise1, . . . , en ∈ V . Then the matrices π(g) =

(aπij(g)

)are unitary matri-

ces. Since for unitary matrices U we have U−1 = U? (conjuge trans-pose), we have aπij(g) = aπji(g

−1).There is a standard scalar product on C[G] defined by

(f1, f2) =1

|G|∑g∈G

f1(g)f2(g).

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Then the orthogonality relations take the form:

Theorem 2.6. For every irreducible representation π choose an in-variant scalar product and an orthonormal basis eπ1 , . . . , edimπ

n andcorresponding matrix coeffecients aπij. Then the functions aπij are mu-tually orthogonal with (

aπij , aπij

)=

1

dimπ.

Proof. This is just a reformulation of the Theorem.

Let (V, π) be an irreducible representation of G and (W, θ) anyfinite-dimensional representation. Denote by W (π) the space spannedby elements Tx for T ∈ Mor(π, θ). It is an invariant subspace of W ,called the isotypic component of θ corresponding to π.

Lemma 2.7. The operator P : W →W defined by

P =dimπ

|G|∑g∈G

χπ(g−1)θ(g),

where χπ(g) = Trπ(g), is a projection onto W (π) along∑[π′]∈G\[π]W (π′). In particular,

W =⊕

[π]∈G

W (π)

andθ∣∣W (π)

∼ π⊕nπ ,

where nπ is the multiplicity of π in θ.

Proof. For now, see Theorem 8, page 2.6 in [?].

2.3 Decomposition of the regular representationWe want to decompose the left regular representation λ into irre-ducibles.

It is sometimes more convenient to work with ρ, the right regularrepresentation. Note that the map C[G]→ C[G] defined by δg 7→ δg−1

defines an equivalence (check!)1.The followig theorem was already proved using the density theorem,

but we now get a more explicit proof.

1Recall that the right regular representation is defined by (f · h)(k) = f(hk)

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Theorem 2.8. Every irreducible representation π of G has multiplicitydimπ in the regular representation, so

λ ∼ ρ ∼⊕

[π]∈G

π⊕ dimπ.

Proof. Fix a basis v1, . . . , vn of Vπ. In matrix form the identity ρ(gh) =ρ(g)ρ(h) takes the form

aπij(gh) =∑k

aπik(g)aπkj(h).

Or equivalently,ρ(h)aπij =

∑k

aπkj(h)aπik.

Therefore, for every i, the map

Vπ → C[G], ej 7→ aπij (1 ≤ j ≤ n)

is an intertwiner between π and ρ.Assume now that V ⊂ C[G] is a ρ-invariant subspace such that

ρ∣∣V∼ π. Choose a basis f1, . . . , fn on V such that

ρ(h)fj =∑k

aπkj(h)fk,

sofj(gh) =

∑k

aπkj(h)fk(g).

Thusfj =

∑k

fk(e)aπkj

by setting g = e.We thus conclude that the isotypic component corresponding to π

equals Spanaπijni,j=1. The functions aπij are linearly independent bythe orthogonality relations, hence their span have dimension n2, andthe multiplicity of π in ρ is n.

Corollary 2.9. There are only finitely many equivalence classes ofrepresentations of G, and∑

[π]∈G

(dimπ)2 = |G|.

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3 Character theoryLet (π, V ) be a representation.

The function χπ : G→ C defined by Trπ(g) is called the characterof π.

We say that a function f : G→ C is central if f(hgh−1) = f(g) forall h, g ∈ G. Note that characters are central, by standard propertiesof traces. Also note that χπ(e) = dimπ.

Theorem 3.1. The characters χπ (for [π] ∈ G) form an orthonormalbasis for the space of central functions on G.

Proof. We have

χπ(g) =

dimπ∑i=1

aπii(g),

and the orthogonality relations show that χπ | [π] ∈ G is an or-thonormal system:

(χπ, χπ) =

dimπ∑ij

(aπii, aπjj) =

dimπ∑i=1

1

dimπ= 1.

Thus we need to show that the characters span the space of centralfunctions. Consider the projection P from C[G] to the space of centralfunctions defined by

f 7→ (Pf)(g) =1

|G|∑h∈G

f(hgh−1).

It is obviously a projection. Now consider P (aij)(g) (we skip the upperindex π). Then this is equal to

1

|G|∑h∈G

aij(hgh−1) =

1

|G|∑h∈G

∑k

∑l

aik(h)akl(g)alj(h−1).

But after using that the orthogonality relations, this simplifies to

1

dimπδijχπ(g).

But the aπij constitute a basis of C[G], by the proof of Theorem ??.Thus image(P ) = Spanχπ.

The dimension of the space of central functions coincides with thenumber of conjugacy classes c(G) in G. We thus have:

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Corollary 3.2. Let c(G) be the number of conjugacy classes in G. Wehave

|G| = c(G).

Corollary 3.3. Two finite-dimensional representations π and θ areequivalent if and only if χπ = χθ.

Proof. You can read off the multiplicities from the character.

Corollary 3.4. For any [π] ∈ G, the multiplicity of π in the regularrepresentation λ equals dimπ.

Proof. We are interested in (χλ, χπ). But note that

χλ(g) =∑h∈G

Tr(δh 7→ δhg) =

0 if g 6= e

|G| if g = e.

Then (χλ, χπ) = 1|G|χλ(e) · χπ(e) = dimπ.

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4 The Frobenius determinant problemLet G be a finite group and C[xg | g ∈ G] the algebra on generatorsindexed by g ∈ G. Then one can form the following determinant:

PG = det ((xgh)gh) .

It is the determinant of the multiplication table of G. The problem isto decompose PG into irreducible polynomials.

It is conventient to instead work with PG = det((xgh−1)gh

). Thishas the same determinant as PG up to multiplication by ±1 since thisjust corresponds to permuting some of rows of the multiplication table.

Now here comes the key observation: the matrix (xgh−1) is thematrix of the operator ∑

g∈Gxgλ(g)

in the basis δg ∈ C[G]. Here you are supposed to view elements ofC[xg | g ∈ G] as functions on G.

That this operator actually has matrix (xgh−1) can be seen by ap-plying the operator to basis elements:∑

g∈Gxgλ(g)

(δh) =∑g∈G

xgδgh

by definition of the left regular action. Reparametrizing this last sumby setting g′ = gh we get∑

g∈Gxgδgh =

∑g′=gh

xgh−1δg,

as wanted.Now, the regular representation decomposes as

λ ∼⊕

[π]∈G

π⊕ dimπ

Hence the determinant decomposes as well (since the operator is G-invariant) and we get

PG =∏

[π]∈G

det

∑g∈G

xgπ(g)

dimπ

The question is now of course if we can decompose further. Theanswer is no:

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Theorem 4.1 (Frobenius). The polynomials det(∑

g∈G xgπ(g))

([π] ∈ G) are irreducible and no two of them are associates.

We begin with a lemma:

Lemma 4.2. The polynomial p = det(xij) is irreducible.

Proof. Let X = (xij). Note that since p is linear in each variable, p isa sum of monomials with no square terms. Suppose for contradictionthat detX = fg and that any determinant of a (n−1)×(n−1) matrixis irreducible (by induction).

Then one of f, g must be independent of xnn, say f . Then g islinear in xnn, hence can be written as g = xnna + b, where a, b arepolynomials in which xnn does not appear. Hence

detX = fg = xnnaf + bf.

Now let Y be the square matrix obtained from X by removing the lastrow and column. By induction detY is irreducible, and have coefficientxnn in the expansion of detX. Hence af = detY . Hence either a = 1and f = detY or a = detY and f = 1. The last option would showthat detX is irreducible. So assume that a = 1 and f = detY . ThendetX = xnn detY + bdetY .

This leads to a contradiction: let M be the matrix (δi,n−i). ThendetY = 0 but detX 6= 0.

Proof of theorem: Suppose that Pπ = p1p2. The polynomials p1, p2 arehomogeneous. Fix a basis e1, . . . , en in Vπ. Consider the correspondingmatrix units mij defined by

mijek = δjkei.

Since π(C[G]) = End(Vπ) by the density theorem,we can find aij(g) ∈C such that

mij =∑g∈G

αij(g)π(g).

Consider the ring map C[xg | g ∈ G]→ C[xij ] defined by

xg 7→∑ij

αi,j(g)xij .

Under this map the operator PG becomes∑g∈G

xgπ(g) 7→∑g∈G

∑i,j

αij(g)xijπ(g) =∑i,j

mijxij = (xij).

17

Since the determinant is functorial with respect to ring maps, adecomposition PG = p1p2 would induce a decomposition of det(xij),which was just proved to be impossible.

We still have to check that no two of the determinants are equal(up to associates). But note that Pπ

∣∣xg=0,g 6=e = xdimπ

e . So we canrecover the dimension of π from the determinant. Also note that if wefix g ∈ G, then

Pπ∣∣xe=1,xh=0 for h6=g = det(1 + xgπ(g)) = 1 + Trπ(g)xg + h.o.t.

so we can recover the trace as well. But a representation is determinedby its character, so if the polynomials are equal, then the representa-tions are equivalent. The converse is similar.

4.1 Two constructions on representationsRecall that the tensor product of V and W have a basis vi⊗vj . HencedimC V ⊗W = dimC V · dimCW .

Assume (V, π) is a finite-dimensional representation and considerV ∗ = HomC(V,C). Then we define the contragradient representa-tion πc to be the representation of the dual space defined by

(π(g)f)(v) = f(π(g−1)v).

If v1, . . . , vn is a basis for V and π(g) = (aij(g)), then in this basis

πc(g) = (aji(g−1))ij .

In particular, the character χπc(g) = χπ(g−1) = χπ(g).Also, we can define the tensor product of two representations

π and θ by (π ⊗ θ)(g)(v ⊗ w) = π(g)v ⊗ θ(g)w.Also: Hom(U,W ⊗ V ∗) = Hom(U ⊗ V,W ).Frobenius reciprocity:

Mor(π, η ⊗ θc) ' Mor(π ⊗ θ, η).

Denote by R(G) the abelian group generated by classes [π] ∈ G offinite-dimensional representations and relations [π] + [π′] = [π ⊕ π′].Then it is an exercise to show that R(G) is a free abelian group withbasis [π] ∈ G.

Via the tensor product we can form a product on R(G), making itinto a ring. See the Appendix for an explicit example.

18

5 Dimension of irreducible representationsRecall that ∑

[π]∈G

(dimπ)2 = |G|.

In this lecture we will prove the following theorem:

Theorem 5.1. The dimension of any irreducible representation di-vides the number |G/Z(G)|, where Z(G) is the center of G.

Later we will strengthen this and show that Z(G) can be replacedby any normal abelian subgroup of G.

In order to prove this, we need to introduce some results fromcommutative algebra. Let R be a unital commutative ring and S ⊂ Ra unital subring. Then we say that an element a ∈ R is integral overS if it satisfies a monic polynomial an + s1a

n−1 + . . .+ sn−1 = 0 withcoefficients in S.

We say that a complex number integral over Z is an algebraicinteger. An integral domain S is called integrally closed if anyelement in the fraction field of S integral over S is already in S. It isan easy exercise to see that Z is integrally closed (and so is any UFD).

Lemma 5.2. Let S ⊂ R be as above. Then a ∈ R is integral over S ifand only if the subring S[a] is a finitely-generated S-module.

Proof. This is Proposition 5.1 in Atiyah-MacDonald, [?].

It is an exercise to see that the set of elements of R integral overS actually form a ring. Alternatively, consult Corollary 5.3 in Atiyah-MacDonald.

Proposition 5.3. Assume that π is an irreducible representation. Letg ∈ G. Let C(g) be the conjugacy class of G. Then

1. The number χπ(g) is an algebraic integer.

2. The number |C(g)|dimπ χπ(g) is also an algebraic integer.

Proof. i). Since G is finite, we have gn = e for some n. Hence alleigenvalues of g are nth roots of unity which are algebraic integers. Butχπ(g) is the sum of the eigenvalues, and the set of algebraic integersform a subring of C, hence χπ(g) is an algebraic integer as well.

The second part is more difficult. Let p be the characteristic func-tion of C(g), that is the function defined as follows:

p(x) =

1 if x ∈ C(g)

0 otherwise.

19

Then p lies in the center of C[G]. Consider the subring Z[G] ⊂C[G] (the group ring of G). Let R ⊂ Z[G] be the subring of centralfunctions. Then p ∈ R. As R is a finitely generated abelian group, anyelement of R is integral over Z · 1 ⊂ R.

It follows that π(p) is integral over Z · idVπ ⊂ End(Vπ). Since p iscentral, π(p) ∈ EndG(Vπ) = C · idVπ . Hence π(p) = α · idVπ for somecomplex number α. This α is an algebraic integer, since π(p) is integralover Z.

But now

α · dimπ = Tr(α · idVπ )

= Trπ(p)

=∑

h∈C(g)

Trπ(h)

=∑

h∈C(g)

χπ(h)

= |C(g)|χπ(g).

The last equality is because characters are constant on conjugacyclasses. This proves the theorem.

Lemma 5.4. If π is irreducible, then dimπ | |G|.

Proof. Recall that(χπ, χπ) = 1.

This can be rewritten as

|G|dimπ

=∑g∈G

χπ(g)χπ(g−1)

dimπ.

Now let C(G) denote the set of conjugacy classes in G. C is then apartition of G. Then the above sum can be rewritten as∑

C∈C(G)

∑h∈C

χπ(h)χπ(h−1)

dimπ=

∑C∈C(G)

|C|χπ(gC)

dimπχπ(g−1

C ),

where gC is any representative from C. The right-hand-side is a sumof products that we know are algebraic integers. They form a ring,hence the sum is an algebraic integer as well.

But the left hand side is a fraction, so we must have |G|/dimπ ∈ Z,which is equivalent to what was to be proven.

Now we can prove Theorem ??. Recall that we want to show thatdimπ divides |G/Z(G)|.

20

Proof. For every n ∈ N, consider the representation of Gn = G×· · ·×Gon Vπ⊗. . .⊗Vπ defined by πn(g1, . . . , gn) = π1(g1)⊗. . .⊗πn(gn). This isan irreducible representation, since any elementary tensor v1⊗ . . .⊗vncan be sent to any other elementary tensor by the action of Gn.

Now consider the subgroup

Zn = (g1, . . . , gn) ∈ Z(G)n |∏

gi = e.

It is isomorphic to Z(G)n−1. If g ∈ Z(G), then by Schur’s lemma,π(g) is a scalar operator. So if (g1, . . . , gn) ∈ Zn, then π(gi) = αi · 1,hence

πn(g1, . . . , gn) = α1 · . . . · αnidV ⊗n = idV ⊗n

since g1 · · · gn = e. Therefore Zn ⊂ kerπn. It follows that πn inducesa representation of Gn/Zn. Then by the previous lemma

(dimπ)n | |G|n

|Z(G)|n−1

That is: (|G|

dimπ|Z(G)|

)n∈ 1

|Z(G)|Z

for any n. This is true for any n, hence

Z[

|G|dimπ|Z(G)|

]⊂ 1

|Z(G)|Z.

This implies that the left-hand-side is a finite-generated Z-module, soby Lemma ??, the fraction |G|/ dimπ|Z(G)| is an algebraic integer.Hence it is an integer. This proves that dimπ | |G|/|Z(G)|.

21

6 Representation theory of SnIt is easy to understand conjugacy classes in Sn. Every σ ∈ Sn definesa partition of [n] = 1, . . . n into orbits of σ. Let O1, . . . , Om be theseorbits. Assume that they are decreasing in size. Let ni = |Oi|. Thesenumbers ni sum to n.

We say that a partition of n is a decreasing sequence of integersn1 ≥ . . . ≥ nm ≥ 1 such that n =

∑ni.

Thus every σ ∈ Sn gives a partition of n. Two elements of Snare conjugate if and only if they define the same partition. To seethis, note that a permutation is determined by its decomposition intodisjoint cycles. If σ = σ1 . . . σr is a product into disjoint cycles, thenρσρ−1 = ρσ1ρ

−1 . . . ρσrρ−1 is also a decomposition into disjoint cycles.

Hence conjugat elements define the same partition. For the converse,write the partition n = n1 + n2 . . .+ nm as

(1, 2, . . . , n1)(n1 + 1, . . . , nn2) · · ·

Then renaming the numbers correspond to conjugating.Our goal is to produce an irreducible representation for every par-

tition of n.Recall that for any finite group G, we have

C[G] '⊕

[π]∈G

End(Vπ),

the isomorphism sending g to the corresponding endomorphism of V =⊕Vπ. Now one can ask if it is possible to recover the space Vπ from itsmatrix algebra? (essentially, yes)

We need a few definitions. An idempotent e in an algebra A iscalled minimal if e 6= 0 and eAe = Ce.

Lemma 6.1. Let V be a finite-dimensional vector space. Show thate ∈ End(V ) is a minimal idempotent if and only if it is a projectingonto a 1-dimensional subspace Cv ⊂ V . Then we have an isomorphism

End(V )e ' V

of End(V )-modules given by Te 7→ Tv.

Proof. –TO COME–

Therefore finding irreducible representations of G is the same asfinding minimal idempotents in the group algebra C[G]: if e ∈ C[G] isa minimal idempotent, then C[G]e defines an irreducible representation([[WHY IS IT IRREDUCIBLE??]].

Thus, returning to Sn, we want to to construct minimal idempo-tents in C[Sn] for every partition.

22

It is convenient to present partitions as Young diagrams. Givena partition (n1, . . . , nm) of n, we draw a diagram having n1 boxes inthe first row, n2 boxes in the second, and so on. For example, for thepartition (6, 5, 2, 1) of 14, we draw the following diagram:

We can also fill these with numbers: a Young tableau (pluraltableaux ) is a Young diagram filled with numbers i ∈ [n] without rep-etitions. For example:

3 12

Two Young tableaux are said to be of the same shape if they arisefrom the same underlying Young diagram.

Note that the symmetric group act on the set of Young tableaux ofthe same shape λ.

Fix a Young tableau T . Let R(T ) ⊂ Sn be the subgroup of ele-ments of Sn permuting numbers in the rows of T . Let C(T ) be thesame group, but permuting columns instead. Define the following twoelements of C[Sn]:

aT =1

|R(T )|∑

g∈R(T )

g

andbT =

1

|C(T )|∑

g∈C(T )

sgn(g) · g.

Then we define cT = aT bT . The element cT is called a Young sym-metrizer. Our main theorem is this:

Theorem 6.2. 1. For any T , the element cT is (up to a scalar) aminimal idempotent in C[Sn], so it defines an irreducible repre-sentation of Sn.

2. If T1, T2 are two Young tableaux, then the representations corre-sponding to them are equivalent if and only if T1 and T2 have thesame shape.

The modules C[Sn]cT are called Specht modules.

Proof. TOO LONG FOR NOW. WILL PROB. COME LATER

23

6.1 TabloidsA bit more explicitly the Specht modules can be described as follows.

Fix a Young diagram λ. Introduce an equivalence relation on Youngtableaux of shape λ:

T1 ∼ T2

if T1 = r(T2) for some r ∈ R(T2). The equivalence class of a Youngtableau T is called a tabloid (of shape λ) and denoted by T.

Note that Sn act on the set of tabloids. Let (Mλ, Sn) be the corre-sponding permutation representation. Note also that action on the setof tabloids is transitive. Thus the basis of Mλ can be identified withSn/R(T )2.

Thus we get isomorphisms of C[Sn]-modules:

C[Sn]aT ' C[Sn/R(T )] 'Mλ.

The first isomorphism is given by sending gaT to ∂gR(T ) (the deltafunction), and the second is given by sending a coset ∂gR(T ) to g(T ).

Under these isomorphisms, the image of the submoduleC[Sn]bTaT ⊂ C[Sn]aT is spanned by the elements bg(T )g(T ). Inother words, Vλ is spanned by eT ′ := bT ′T ′ by all elements of shapeλ. This gives us a description of Vλ ' VT only in terms of λ.

This is a spanning set. We would like a basis. We say that thattableau T is standard if the numbers in every row and column areincreasing.

Here’s a theorem that we will not prove:

Theorem 6.3. For any Young diagram λ the elements eT for all stan-dard tableaux T of shape λ form a basis in Vλ.

The complication is of course that geT = cg(T ) is a combination ofstandard tableaux!

Example 6.4. Let λ = n, be the trivial partition. Then the corre-sponding Young diagram looks like:

...

For any tableau T of shape λ we get C(T ) = e, since all the columnsare one-element sets. Also, R(T ) = Sn. Hence

cT =1

n!

∑g∈Sn

g.

Then gcT = cT g = cT . Hence C[Sn]cT = CcT , so that πT ∼ ε, thetrivial representation. F

2Every T can be written as σT0 for some fixed T0. But the ambiguity lies in R(T ).

24

Example 6.5. Let λ = (1, . . . , 1) be the partition n = 1 + . . . + 1.Then the corresponding Young diagram looks like

...

Then C(T ) = Sn and R(T ) = e. Therefore

cT =1

n!

∑g∈Sn

sgn(g) · g.

So gcT = cT g = sgn(g)cT . Hence C[Sn] = CcT , and πT is the one-dimensional representation sgn. F

6.2 Characters and the Hook length formulaDenote by πλ the irreducible representation of the symmetric groupcorresponding to the Young diagram λ.

To describe the character of πλ, it is convenient to index the con-jugacy classes in Sn by the number of cycles.

Let I = (i1, i2, . . . , ) consist of non-negative integers such that

∞∑k=1

kik = n.

Denote by cI the conjugacy class in Sn consisting of elements σ whichdecompose into i1 orbits of length 1, i2 orbits of length two and so on.

This correspond to the partition kik + (k − 1)ik−1 + ... = n.Let χπλ(CI) denote the value of the character of πλ on any repre-

sentating of the conjugacy class of CI . Then

Theorem 6.6 (Frobenius character formula). Assume λ =(λ1, . . . , λr). Take any number N ≥ r. Then χπλ(CI) is the coeffi-cient of

∏Ni=1 x

λi+N−ii in the polynomial

∆(x)∏k≥1

(N∑i=1

(xki )

)ik.

where ∆(x) =∏

(xi − xj) is the Vandermonde determinant.

Proof. If time allows a proof will appear here.

25

Now we quote the hook length formula. Let λ be a Young diagramand let (i, j) be the element in position row i and column j. Then thehook length h(i, j) is the number of boxes below and to the right of(i, j) including (i, j) itself.

Using the notation in the previous Theorem, let li = λi+N − i (weput λi = 0 for i > r). Then:

Lemma 6.7. For any i, we have that

li!∏nj=i+1(li − lj)

=

λi∏k=1

h(i, k).

From this we get:

Theorem 6.8. Let λ be partition and let πλ be the corresponding ir-reducible representation of Sn. Then

dimπλ =n!∏

(i,j)∈λ h(i, j).

This is also the number of standard Young tableaux of shape λ.

Prooof later.

Example 6.9. Let

be a Young diagram. Then dimπλ = 3!1·1·1·3 = 2. F

26

7 Induced representationsLet G be a finite group and H ⊂ G a subgroup. Assume (π, V ) is arepresentation of H. We want to construct a representation of G outof this.

We define the induced representation IndGHπ to be the C[G]-module define by

C[G]⊗C[H] V.

It is very useful in constructing representations of larger groupsfrom smaller ones. The idea is that if we know the representations somesubgroup H of a group G, then in nice situations all representations ofG can be found as subrepresentations of IndGH π for various [π] ∈ H.

Remark. This is exactly the concept of extension of scalars from com-mutative algebra (see e.g. Atiyah-MacDonald page 28). Namely letA = C[H] and B = C[G] and let M = V be an A-module. Then we canform the A-module MB = B ⊗AM , which carries a natural B-modulestructure as well: the action of B on MB is given by b(b′⊗x) = bb′⊗x.

We have three equivalent description of the induced representation,some of the descriptions being easier to work with than others.

Theorem 7.1. Let V be a representation of H. The following repre-sentations are isomorphic.

1. C[G]⊗C[H] V where G acts by g · (g′ ⊗ v) = gg′ ⊗ v.2. HomH(C[G], V ) ' f : G → V | f(hg) = hf(g)∀h ∈ H, g ∈ G,

where g acts by (g′ · f)(g) = f(gg′).

3. W =⊕

σ∈G/H Vσ, where each Vσ is a copy of V . Let xσ ∈ Vσ.Then the action of G is defined as follows: Let rσ ∈ σ be arepresentative for the coset σ. Then g can be written as g = rσ′hfor some h ∈ H. Then we define g · xσ = h · xgσ.

Proof. Start with 1). We want to show that this is the same as number3). We will do this in a non-canonical way. The algebra C[H] havea vector space basis eh1 , . . . , ehr where r = |H|. By properties ofcosets, this implies that C[G] have a basis

eh1 , . . . , ehr , er1h1 , . . . , er1hr , . . . , erkh1 , . . . , erkhr

where k = [G : H] and the ri’s are a set of fixed representatives for thecosets G/H. This means that we can write

C[G] =

k⊕i=1

riC[H].

27

This implies that C[G]⊗C[H] V decomposes as

C[G]⊗C[H] V =

(k⊕i=1

riC[H]

)⊗C[H] V '

k⊕i=1

(riC[H]⊗C[H] V

)This is exactly the description in 3), because each riC[H] ⊗C[H] V isisomorphic to V via the map rih ⊗ v 7→ h−1v (we need the inverse inorder for this to be a map of representations). The representation Vis endowed with a left action of H (this is what a representation is),but it also has a right action, defined by v · g := g−1v. In this way, Vbecomes a left-right-H-module.

One can check that the action in 3) and the natural G-action onriC[H]⊗C[H] V are the same.

The equivalence of 1) and 2) can be seen by standard isomorphismsfrom module theory. Namely, note that

HomH(C[G], V ) ' HomH(C[G],C)⊗C[H] V.

We also have an identifaction of representations C[G] withHom(C[G],C) where we let δg be a basis of C[G]. Then the actionof G on C[G] by left-multiplication is identified with the contragra-dient action of G on Hom(C[G],C). But all functions C[G] → C areG-invariant (because f(gx) = gf(x) = f(x) since we are consideringthe trivial action on C). Hence HomH(C[G],C) = C[G]. So the twodescriptions are equal.

We say that the collection Vσ ⊂ W is called a system of im-primitivity of the representation V .

Example 7.2. Let λH be the regular representation of H. Then λHcan be identified with C[H]. In this case we have IndGH λH = λG sinceC[G]⊗C[H] C[H] ' C[G]. F

Here are some basic properties of induction. Most of them followfrom basic properties of tensor products.

1. If N ⊂M ⊂ G are subgroups, then

IndGH IndMN π ∼ IndGN π

2. If π is finitedimensional, then

(IndGH π)∨ ' IndGH π∨.

Here π∨ is the contragradient representation of π.

28

7.1 Frobenius reciprocityLet Rep(G) denote the category of finite-dimensional G-representationswhere G is a finite group and let H ⊂ G be a subgroup of G. Wehave two functors in each direction: we have a functor ResGH sendinga representation to its restriction to H, and we have a functor IndGHin the other direction, sending a representation of H to the inducedrepresentation on G. The theorem of this section is that these twofunctors are adjoints to each other.

But this is a standard fact about tensor products: let S ⊂ R berings (commutative or not) and let V be an S-module andW an R−S-bimodule. In our case a representation W of G is both a C[G]-moduleand a C[H]-module by restriction. Then we always have

HomR(R⊗S V,W ) ' HomS(V,W )

Note that as an S-module we have W = ResGHW .If χ is the character of π (a representation of H), we denote by

IndGH χ the character of the induced representation.Note that it follows by Schur’s lemma and the orthogonality rela-

tions that for any two representations, we have

(χπ1, χπ2

) = dimC Mor(π1, π2).

Therefore, in terms of characters, Frobenius reciprocity can be statedas

(IndGH χπ, χθ) = (χθ,ResGH χθ).

Thus:

Proposition 7.3. Let V be an irreducible representation of H and Wan irreducible representation of G. Then the number of times that Woccurs in IndGH V is equal to the number of times that ResGHW occursin V .

Proof. Recall that if V is an irreducible representation and U is anyrepresentation, then (χV , χU ) is equal to the number of times that Voccurs in U . Now use the above remarks.

7.2 Characters of induced representationsProposition 7.4. Assume G ⊃ H and that (π, V ) is a finite-dimensional representation of H and χ its character. Then the char-acter IndGH χ of IndGH π is given by

(Indχ)(g) =∑

s∈G/H,s−1gs∈H

χ(s−1gs) =1

|H|∑

s∈G|s−1gs∈H

χ(s−1gs),

where we extend χ to G by zero.

29

Proof. Consider the imprimitivity system Wxx∈G/H for the inducedrepresentation θ = IndGH π. Then

χθ(g) = Tr(g)

=∑

x∈G/H,gx=x

Tr(θ(g)∣∣Wx

)

=∑

s∈G/H,s−1gs∈H

Tr(θ(g)∣∣Ws

)

=∑

s∈G/H,s−1gs∈H

Tr(θ(s−1)θ(g)θ(s)∣∣We

)

=∑

s∈G/H,s−1gs∈H

Tr(π(s−1gs))

=∑

s∈G/H,s−1gs∈H

χ(s−1gs).

The second equality follows by noting that χ(s−1gs) is constant on leftcosets as s varies.

This makes it easy to compute characters of induced representa-tions. In practice one often induces from a normal subgroup containingfew conjugacy classes, so that many of the terms in the sum are zero.

30

8 The Frobenius subgroup theoremIn this section we will prove a theorem of group theory, the proofof which relies heavily upon representation theory. To this day, no“elementary” proof is known.

Theorem 8.1. Assume a finite group G acts transitively on a set Xsuch that every g 6= e has at most one fixed point. Then there exists aunique normal subgroup K of G acting freely transitively on X.

Proof. Put

K∆= e ∪ g ∈ G | g has no fixed points .

We have to show that K is a normal subgroup and that it acts freelytransitively on X. Once this is done, we are finished, since K is clearlythe largest possible subset ofG satisfying the conclusion of the theorem.

To show that K acts freely transitively on X, it suffices to showthat #K = #X. Let x ∈ X, and consider the stabilizer

H = Gx = g ∈ G | g · x = x.

If g 6∈ H, by definition g · x 6= x. Also note that

H ∩ gHg−1 = Gx ∩Ggx = e,

since otherwise h in the intersection would act trivially on both x andgx, but we assumed that every element acted with at most one fixpoint.

Hence we can write G as a disjoint union as follows:

G = K∐∐

y∈XGy\e

= K∐ ∐

[g]∈G/H

gHg−1\e

.

Therefore#G = #K +

#G

#H(#H − 1)

so that #K = #G/#H = #X.It remains to show that K is a normal subgroup. We will do this

by showing that it is the kernel of some representation of G.For an irreducible representation π of H, consider the function χπ

on G defined by

χπ = IndGH χπ − (dimπ) IndGH χεH + (dimπ)χεG

= IndGH χπ − (dimπ) IndGH 1 + (dimπ)1.

31

Recall that in general, if f : H → C is any class function, we caninduce a new class function on G by the formula

(IndGH f)(h) =∑

[g]∈G/H,g−1hg∈H

f(s−1gs).

Here the sum chooses one representative from each coset. Now onesees easily that

χπ(g) =

dimπ if g ∈ Kχπ(h) if g is conjugate to h ∈ H

In the first case we get 0−0+dimπ in the definition, and in the secondcase we get IndGH χπ(g)−dimπ+dimπ. That this completely describesχπ(g) is clear because of our description of G as a disjoint union above.

Using this decomposition, we can calculate (χπ, χπ):

(χπ, χπ) =1

|G|∑g∈G|χπ(g)|2 =

1

|G||G||H|

dimπ2+1

|G||G||H|

∑h∈H\e

|χπ(h)|2

But this last expression is equal to (chiπ, χπ) = 1.As χπ is a linear combination of characters with integer coefficients,

we must have χπ = ±χπ for some irreducible representation π.3. Butχπ(e) = dimπ > 0, so that actually χπ = χπ.

Note that as χπ∣∣H

= χπ, we have ResGH π ∼ π.Now we claim that

K =⋂

[π]∈H

ker π.

This implies that K is a normal subgroup. This would finishing theproof.

First we look at ⊆: Let g ∈ K. Then χπ(g) = dimπ = dim π.But π(g)n = 1 for some n ≥ 1. Hence all the eigenvalues of π(g) areroots of unity. The matrix π(g) is diagonalizable, hence the sum ofdim π roots of unity is equal to dim π. This is only possible if all theeigenvalues are 1 by the triangle inequality, i.e. that π(g) = id. Thusg ∈ ker π.

Now look at ⊇. Let g ∈ G\K. Then g is conjugate to an elementh ∈ H\e. Then we can find some irreducible representation [π] ∈ Hsuch that π(h) 6= 1. Hence π(g) 6= 1.

This finishes the proof.

3To see this, note that the norm of χπ is a sum of squares of integers, summing to 1.

32

Example 8.2. Let k be a finite field and let G be the translationgroup of A1, or in other words the “ax+b”-group. It is isomorphic tothe matrix group

G '

a b

0 1

| a ∈ k×, b ∈ k .

Then G satisfies the conditions of the theorem (we say that G is aFrobenius group). Then the group K in the theorem is the group oftranslations:

K =

1 b

0 1

| b ∈ k .

F

33

9 The theorems of Artin and Brauer andfields of definitionTheorem 9.1 (Artin). Let G be a finite group and C a collection ofsubgroups of G. Then the following two conditions are equivalent:

1. Any character of G is a rational combination of characters ofrepresentations induced from subgroups in C .

2. Any g ∈ G is conjugate to an element in a subgroup in C .

Proof. The proof is “surprisingly simple”.We first prove 1⇒ 2. Recall that the characters span the space of

central functions on G.At the same time, the characters induced from subgroups in C are

zero on elements that are not conjugate to elements of subgroups inC , by Proposition ??. But any non-identity g ∈ G have a characterwith non-zero value at g (take for example the induced regular repre-sentation). Thus g must meet some conjugacy class in C .

Conversely, denote by RC(G) the space of central functions on Gwith the usual scalar product.

Consider the map

T :⊕H∈C

RC(H)→ RC(G)

T((f)H∈C

)=∑H∈C

IndGH fH .

Recall that by Frobenius reciprocity, the functor ResGH is adjoint toIndGH . Therefore the adjoint map of T is given by

T ∗ : RC(G)→⊕H∈C

RC(H)

sending f to (ResGH f)H∈C .Now suppose ResGH f = 0 for all H ∈ C and some f ∈ RC(G). But

as above, since any g ∈ G is conjugate to an element from some H, itfollows that f is zero on all conjugacy classes, hence f is zero. So themap T ∗ is injective, hence T is surjective.4

This means that we can find a basis in RC(G) consisting of charac-ters χ1, . . . , χn of representations induced from subgroups in H ∈ C .The characters χ′1, . . . , χ′n of irreducible representations of G form abasis in RC(G) by Prop .... Therefore the transition matrix A fromχ1, . . . , χn to χ′1, . . . , χ′n has integer entries. Hence A−1 has ratio-nal entries.

4Quote Sergey: “we didn’t do any real work”.

34

Therefore any character induced from an irreducible representationcan be written as a Q-linear combination of the χi’s. Therefore thisholds for all characters.

We can take C to be the collection of cyclic subgroups of G.

Corollary 9.2. Any character can be written as a Q-linear combina-tion with rational coefficients of characters induced from cyclic sub-groups.

Proof. Any g ∈ G is contained in some cyclic subgroup, say 〈g〉.

The following is more difficult to prove (we won’t):

Theorem 9.3 (Brauer). For any finite group G, the characters ofG can be written as a linear combination with integer coefficients ofcharacters IndGH χπ of one-dimensional representations π of subgroupsH ⊂ G.

In fact we can let H range through products of cyclic and p-subgroups.

For a proof, see for example [?].

9.1 Field of definitionLet π : G → GL(V ) be a representation over the complex numbers.Let K ⊂ C. We say that π is defined over K if there exists a vectorspace V ′ over K and a representation π′ : G→ GL(V ′) such that π isequivalent to π′C, where

π′C(g) = 1⊗ π(g) : C⊗K V ′ → C⊗K V ′.

Example 9.4. Let G = Z/nZ. Representations of G are characters,all of the form e

2πin . Hence any representation of G is defined over

Q(ζn). F

Example 9.5. If G = Sn (a much more complicated group), everyrepresentation is in fact defined over Q. F

As long as the field K is of characteristic zero, many of the resultsare still true with the same proofs. The main difference is that if πis an irreducible representation, then End(π) forms a division algebraover K, instead of being just equal to K itself. There are not manydivision algebras of characteristic zero, so this is not too bad.

Many results remain true, however. Maschke’s theorem (that is,proving complete reducibility) goes through for any field of charac-teristic zero (or for any group of order not dividing the characteris-tic). We also have a weaker form of Schur’s lemma: if π and π′ are

35

non-equivalent irreducible representations, then Mor(π, π′) = 0 and(χπ′ , χπ) = 0.

Note that χπ′C = χπ′ . This implies that disjoint irreducibles staydisjoint when complexifying. What can happen, of course, is thatirreducibles split into disjoint representations.

If G is a finite group, then the exponent of G is the least commonmultiple of orders of all elements in G.

Theorem 9.6 (Brauer). Let G be a finite group and let m be theexponent of G. Then any complex representation of G is defined overQ(ζn), where ζn is a primitive n-th root of unity.

Proof. Let π1, . . . , πk be representatives of the equivalence classes ofirreducible representations of G over Q(ζn).

Note that any one-dimensional representation of a subgroup of Gis defined over Q(ζn). Hence the induction of any such representationis defined over Q(ζn).

Now let π be any finite-dimensional complex representation repre-sentation of G. It follows by Brauer’s Theorem ?? that

χπ =

k∑i=1

miχπiC =

k∑i=1

miχπi

for some mi ∈ Z, since each IndGH χ is a Z-linear combination of theπi’s.

Assume in addition that π is irreducible. Then

1 = (χπ, χπ) =∑i,j

(χπi , χπj )

=∑i

m2i (χπi , χπi).

This is only possible if χπ = χπi for some i, hence π ∼ (πi)C.So π is defined over Q(ζn). Thus the theorem is true for irreduciblerepresentations, and so true for every representation.

36

10 Mackey’s irreducibility criterionWhen is an induced representation irreducible? In practice, this isalmost never the case.

Example 10.1. Let H ⊂ G be a subgroup. Then IndGH εH = C[G/H],which is never irreducible. F

We need a unpleasant lemma.Let G be a finite group and let H,K be two subgroups and suppose

π is a representation of H. Define a subgroup Hg ⊂ H by

Hg = gHg−1 ∩K, g ∈ G.

Then, for any g ∈ G, define a representation πg of Hg by

πg(h)∆= π(g−1hg).

Lemma 10.2. We have

ResGK IndGH π ∼⊕

g∈K\G/H

IndKHg πg,

where K\G/H = G/ ∼, where g1 ∼ g2 if g1 = kg2h with k ∈ K andh ∈ H.

Proof. To come.

Theorem 10.3 (Mackey). Assume π is an irreducible representationof H ⊂ G. Then IndGH π is irreducible if and only if the representationsResHHg π and πg = π(g−1 − g) are disjoint for all g ∈ G\H.

Proof. With the lemma in hand, the proof is quite short.Let χ = χπ. Then we want to compute (IndGH χπ, IndGH χπ). By

Frobenius reciprocity, this is the same as (ResGH IndGH χ, χ). Hence, bythe lemma, this is (using Frobenius reciprocity again)∑

g∈H\G/H

(IndHHg χπg , χ) =∑

g∈H\G/H

(χπg ,ResHHg χ).

Thus π is irreducible if and only if (χπg , χ∣∣Hg

) = 0 for all g 6∈ H.

37

11 Induction from normal subgroupsWe study the following question: let G be a finite group and H ⊂ G anormal subgroup. Can we describe the representations of G in termsof those of H and those of G/H?

Assume (W, θ) is an irreducible representation of G. Let (V, π)be an irreducible representation contained in ResGH θ. Then, for everyg ∈ G, ResGH θ also contains the representation πg on H defined by

πg(h) = π(g−1hg).

Indeed, if T : V → W intertwines π with Res θ (meaning thatTπ(g) = θ(g)T for all g ∈ H), then θ(g)T intertwines πg with θ:

θ(g)Tπg(h) = θ(g)Tπ(g−1hg) = θ(g)θ(g−1hg)T = θ(h)θ(g)T.

Furthermore, if we consider the isotypic components of W (πg) ofRes θ corresponding to πg, then θ(g1)W (πg2) = W (πg1g2).

Thus the space ∑g∈G

W (πg)

is invariant underG, hence it coincides withW , sinceW was assummedto be irreducible.

In other words, Res θ decomposes into copies of πg, g ∈ G.We can define an action of G on H by

[π] 7→ [πg].

What we have shown, is that any irreducible representation Wθ ofG defines a G-orbit S ⊂ H and that

W =⊕x∈S

W (x), θ(g)W (x)W (gx).

For x ∈ H, denote by I(x) ⊂ G, the stabilizer of x in G. It is calledthe inertia subgroup of x. Note that the action of H on H is trivial,so that we always have H ⊂ I(x).

Thenθ ∼ IndGI(x)

(ResGI(x) θ

∣∣∣W (x)

).

(why??)What can be say about ResGI(x) θ? In other words, assume we take

[π] ∈ H. ThenI = I([π]) = g ∈ G | πg ∼ π.

Can we describe all irreducible representations η of I such that ResIH ηis isotypic to π?

38

It can be shown that all such representations of I can be describedin terms of projective irreducible representations of I/H.

In general, a projective representation of a group K on a vectorspace V is a homomorphism K → PGL(V )

∆= GL(V )/C∗ · idV .

Why do projective representations appear? We have that πg ∼ πfor all g ∈ I. Thus there exists an invertible Tg ∈ GL(Vπ) such that

π(g−1hg) = T−1g π(h)Tg

for all h ∈ H. By irreducibility, Tg is unique up to a scalar factor. Thisimplies that Tg1g2

= c(g1, g2)Tg1Tg2

for some c(g1, g2) ∈ C∗. There theoperators Tg define a projective representation of I on V .

In some cases projective representations do not appear. We willmake the following assumptions: H is abelian, so π is a character ofH. Also, π : H → GL(C) extends to a homomorphism π : I → C∗.

We want to describe all irreducible representations of I such thatResIH σ is isotypic to π.

Assume σ is such a representation. Then consider σ = π ⊗ σ(recall that by assumption, π is a character, so conjugation makessense). Then σ is a representation on the same space given byσ(g) = π(g)−1σ(g). Then σ is irreducible and H ⊂ ker σ. So σ isan irreducible representation of I([π])/H.

Conversely, if σ is an irreducible representation of I/H, then wecan define σ by σ = π ⊗ σ.

We have almost proved:

Theorem 11.1. Let G be a finite group and H ⊂ G a normal abeliansubgroup. Assume that every character χ ∈ H extends to a homomor-phism χ : I(χ)→ C∗, where

I(χ) = g ∈ G | χg = χ.

Then the irreducible representations of G can be described as follows:for every G-orbit in H, fix a representative χ. Then for every [σ] ∈I(χ)/H, define IndGI(x)(χ⊗ σ).Therefore, as a set, G can be identified with⊔

[χ]∈H/G

I(χ)/H,

where the [χ] ranges over representatives of G-orbits in H.

Example 11.2. Consider a finite field Fp with p prime. Consider thecorresponding ax+ b group:

G =

a b

0 1

| a ∈ F∗p, b ∈ Fp

.

39

Let H be the subgroup H =

1 b

0 1

| b ∈ Fp

' Z/p. Then H =

Z/p consists of characters χ0, . . . , χp−1 where

χk(b) = e2πip kb.

We have

χ

a b

0 1

= χ(a−1−).

The stabilizer of χ0 is G. The stabilizers of χk for k = 1, . . . , p− 1 aretrivial in G/H, and G acts transitively on χ1, . . . , χp−1.

The assumptions of the theorem are satisfied, as χ0 extends to Gby χ0(g) = 1.

Consider the representatives of the two G-orbits on H:χ0: Here I(χ0) = G. We have G/H ' F×p . This leasds to |F×p | =

p− 1 one-dimensional representations. Namely,

G 3

a b

0 1

7→ χ(a)

for χ ∈ F×p ' Z/(p− 1).χ1: Here (χ1) = H. Then we get one irreducible p− 1-dimensional

representation IndGH χ1. F

11.1 DivisibilityTheorem 11.3. Let G be a finite group, H ⊂ G a normal abeliansubgroup, and π an irreducible representation of G. Then dimπ divides|G|/|H|.

Proof. Recall that we already proved this for H = Z(G). For generalH, we know that π ' IndGI σ for some I ⊃ H and ResIH σ is isotypicto χ ∈ H.

Consider the groups I = σ(I) and H = σ(H) = χ(h) ·1 | h ∈ H.Then H ⊂ Z(I) and the representation of I on Vσ is irreducible.

Hencedimσ | |I/H| = |I/H · kerσ|,

sodimσ | |I/H|.

Hencedimπ = |G/I|dimσ | |G/I||G/H| = |G/H|.

40

12 Part II - Compact groupsWhen one wants to develop a representation theory for infinite groups,one has to restrict the scope somewhat. Infinite groups are too wildin general, and they can have quite bad representation theory. Forexample, Z acting on C2 by (x, y) 7→ (x + ny, y) has the y-axis asan irreducible subspace, but the representation does not split, becausethere is no complementary invariant subspace. Thus the property ofcomplete reducibility fails.

We want to at least restrict to groups with a topology. But whatmade wonders for the theory of finite groups was the presence of anaveraging operator. If we restrict to compact groups, we will see thatwe can mimic this operator in this setting as well.

Recall that a topological group is a group G equipped with atopology such that the group law and the inverse are continous maps.

Here is a key fact:

Proposition 12.1 (Haar, von Neumann, André Weil). For any locallycompact group G there exists a nonzero left-invariant Radon measureon G which is unique up to scalar. We call such a measure a Haarmeasure.

Recall that a Radon measure µ on a locally compact space is aBorel measure which is outer regular, inner regular on open subsetsand µ(K) <∞ for all compacts K.

To be given a Radon measure on a locally compact space X isequivalent to be given a linear functional F : Cc(X) → C such thatF (f) ≥ 0 if f ≥ 0 (here Cc(X) is the space of compactly supportedfunctions on X). Thus we can think of a Radon measure as equivalentto having a way to integrate functions.

If G is compact, we usually normalize so that µ(G) = 1. This makesthe Haar measure unique.

Example 12.2. Let G = (R,+). The the Haar measure is the usualLebesgue measure. F

There are also right-invariant measures. If the left- and right-invariant Haar measures coincide, then we say that the group G isunimodular.

Lemma 12.3. Any compact group is unimodular.

Proof. Let µl be the left-invariant measure. Then the measure µ′l(U) =µl(Ug) is left-invariant as well. Hence µ′l = αµl for some α ∈ R. But1 = µ′l(G) = αµl(G) = α. Hence µ′l = µ. Thus µl(Ug) = µ′l(U) =µl(U), so µl is right-invariant as well.

41

Many of the results for finite groups carry over to compact groups:the proofs are the same except expressions like 1

|G| are replaced by∫Gfdµ.A representation of a compact group G is a continous homomor-

phism π : G→ GL(V ).Here are some results that holds for compact groups as well (with

the same proofs):

• Maschke’s theorem: Any finite-dimensional representation ofG is completely reducible.

• Any finite-dimensional representation is unitarizable.

• The orthogonality relations. In particular, two finite-dimensionalrepresentations of a G are equivalent if and only if the characterscoincide.

The notion of regular representation is however more complicated.We are forced to (briefly) introduce infinite-dimensional spaces.

A unitary representation of a topological group G on a Hilbertspace H is a homomorphism π : G → U(H), continous in the strongoperator topology, meaning the following: if gi → g in G, thenπ(gi)ζ → π(g)ζ for all ζ ∈ H. This can be rephrased as saying thatthe map G→ H given by g 7→ π(g)ζ is continous for all ζ ∈ H.

Example 12.4. Let G be a locally compact group with fixed Haarmeasure. Consider L2(G) (the vector space of square integrable func-tions on G) with respect to the given measure. Define

G // U(L2(G))

g // (λ(g)f)(h) = f(g−1h).

Then λ is continous in the above sense.We call λ the regular representation of the compact group

G. F

Theorem 12.5 (Peter-Weyl). Let G be a compact group and π : G→U(H) a unitary representation. Then π decomposes into a direct sum(in the sense of Hilbert spaces) of finite-dimensional irreducible repre-sentations.

Proof. A sketch was given in the lecture.

Corollary 12.6. For any compact group G, the intersection of allkernels of irreducible representations on G is trivial.

Proof. The regular representation decomposes into a sum of irre-ducibles, and is injective.

42

Denote by C[G] the linear span of matrix coefficients of finite di-mensional representations of G. This is an algebra under pointwisemultiplication: the product of two matrix coefficients (from represen-tations π1, π2, say) is a matrix coefficient in π1 ⊗ π2. This algebra isclosed under complex conjugation, since the matrix coefficient of thedual representation is given by π(g−1)ji. Thus C[G] is a C∗-algebra.

Corollary 12.7. The algebra C[G] is norm-dense in C(G).

Proof. Recall the Stone-Weierstrass theorem: if A is a subalgebra ofC(G) that separates points, then A is norm-dense in C(G). So letg1, g2 ∈ G. By Theorem ??, there exist non-trivial representations ofG. Then [[details later]]

The algebra C[G] is called the algebra of regular/polynomialfunctions on G.

Theorem 12.8 (Peter-Weyl). For every compact group G, we have

λ ∼⊕

[π]∈G

πdimπi ,

where λ : G→ U(L2(G)) is the regular representation.

Just as in the case of finite groups, it is shown that the charactersof finite-dimensional representations form an orthonormal basis in thespace

f ∈ L2(G) | f(g − g−1) = f(−)∀g ∈ G ⊆ L2(G).

In fact, for any compact group G, the characters of finite-dimensional represetantions of G span a dense subspace in the spaceof continous central functions on G.

Theorem 12.9. Suppose G is a compact subgroup of the unitary groupU(n) ⊆ Rn×n. Then C[G] is the unital algebra generated by the matrixcoefficients aij and det−1.

Proof. LATER

43

13 Representation theory of compact LiegroupsA Lie group is a smooth manifold G equipped with smooth mapsµ : G×G→ G and ι : G→ G making (G,µ, ι) into a group.

Equivalently, a Lie group is a group object in the category of smoothmanifolds.

Example 13.1. Let V be a finite-dimensional vector space. ThenGL(V ) is a Lie group. The smooth structure is inherited fromEnd(V ) ' RN , since GL(V ) is an open subset of End(V ). F

Before giving further examples, we will develop some general re-sults.

13.1 The exponential mapWe recall some results from manifold theory.

Let M be a smooth manifold and let p ∈ M . The tangent spaceTpM can be equivalently defined as

• Equivalence classes of smooth curves passing through p: let γi :(ai, bi)→M , with ai < 0 < bi and γi(0) = p be two such curves(i = 1, 2). Then γ1 ∼ γ2 if and only if

(x γ1)′(0) = (x γ2)′(0)

for some/any coordinate system (x, U) around p. We denote theequivalence class by γ′(0), or

• the more algebraic definition: the space of derivations of C∞(M)at p. This is by definition a linear map L : C∞(M) → C suchthat L(fg) = f(p)L(g) + g(p)L(f).

The relation between these two definitions is that the derivation Ldefined by γ′(0) is

L(f) =∂

∂f(f γ)

∣∣∣∣t=0

.

In local coordinates (x, U), any tangent vector can be written as∑ai(p)

∂

∂xi

∣∣∣∣p

.

Recall that a smooth vector field is a smooth section of the tangentbundle. The space of vector fields can be identified with the space ofderivations of C∞(M), i.e. as linear maps L : C∞(M) → C∞(M)satisfying the Leibniz rule.

44

Given a vector field X, an integral curve of X is a smooth curveγ : (a, b) → M such that γ′(t) = Xγ(t) for all t ∈ (a, b). It is astandard fact of manifold theory that for any p ∈ M , there exists aunique maximal integral curve passing through p.

Recall also: if f : M → N is a smooth map, then the differentialat p is the map dpf : TpM → Tf(p)N given by γ′(0) 7→ (f γ)′(0).

Now let G be a Lie group. For each g ∈ G, we have diffeomorphismlg : G → G defined by left-multiplication: h 7→ gh. This induces amap on vector fields:

(lg)∗(X)p = (dg−1(p)lg)(Xg−1p).

We say that a vector field X on a Lie group G is left-invariant if(lg)∗(X) = X for all g ∈ G. Such a vector field is determined by itsvalue at e ∈ G: Xg = (delg)(Xe).

Thus we get a one to one correspondence between left-invariantvector fields and v ∈ TeG.

Proposition 13.2. Let G be a Lie group, X ∈ TeG and αX : (a, b)→G be the maximal integral curve of the corresponding left-invariantvector field such that αX(0) = e.

Then (a, b) = R and αX(s+ t) = αX(s)αX(t) for all s, t ∈ R.

A 1-parameter subgroup of G is a smooth map γ : R→ G thatis a group homomorphism.

Corollary 13.3. The map X 7→ αX is a bijection between TeG andone-parameter subgroups of G.

Proof. Assume γ : R → G is a one-parameter subgroup of G. PutX = γ′(0) ∈ TeG. Then γ = αX :

γ′(t) = (deγ(t))(γ′(0))

as γ(t+ s) = γ(t)γ(s). [[why does it follow??]]

Remark. In general γ(R) is not a closed subgroup of G. Take forexample R → S1 × S1 given by t 7→ (eit, eiθt) for some irrational θ.Then the image is dense in S1 × S1.

We define the exponential map as exp : TeG → G given byX 7→ αX(1).

Note that for any s ∈ R

∂

∂tαX(st) = sα′X(st) = sXαX(st).

Thus t 7→ αX(st) is the integral curve of the vector field correspondingto sX, hence αX(st) = αsX(t).

45

Therefore exp(tX) = αtX(1) = αX(t). So by the corrollary, all1-parameter subgroups of G have the form t 7→ exp(tX).

We have∂

∂texp(tX)

∣∣∣∣t=0

= X.

In other words, if we identify T0(TeG) with TeG, we conclude thatd0 exp = idTeG. Thus exp gives us a diffeomorphism between a neig-bourhood of 0 ∈ TeG and a neighbourhood of the identity e ∈ G.

That exp is smooth is a standard result on integral curves (see e.g.Spivak).

Theorem 13.4. The exponential map is the unique smooth map exp :TeG→ G such that exp((s+ t)X) = exp(sX) exp(tX) and

∂

∂texp(tX)

∣∣∣∣t=0

= X

In general, the exponential map is neither injective nor surjective,even for connected groups.

Example 13.5. Let G = GL(V ). As GL(V ) is a open subset ofEnd(V ), the tangent space at every M ∈ GL(V ) is identified withEnd(V ) for all M ∈ GL(V ). Then exp : End(V )→ GL(V ) is given by

X 7→ eX =

∞∑n=0

Xn

n!.

F

A homomorphism of Lie groups is a homomorphism π : G→ Hwhich is also smooth.

Theorem 13.6. Assume that π : G → H is a Lie group homomor-phism. Then the diagram

Gπ // H

TeG

exp

OO

deπ// TeH

exp

OO

is commutative.

Example 13.7. Let G = GL(V ). Consider the map det : GL(V ) →K× = GL(K). Then de det = trace. Then the theorem says thatdet(eA) = etrace(A). F

46

13.2 The Lie algebra of a Lie groupRecall that for vector field X,Y on a manifold M , there is a bracketoperation [X,Y ]. If LX , LY are the corresponding derivations, then[X,Y ] is defined as the vector field corresponding to the derivationL[X,Y ] = LXLY − LY LX .5 If X and Y are given in local coordinatesas

X =∑

f i∂

∂xi, Y =

∑gi

∂

∂xi,

respectively, then

[X,Y ] =∑i,j

(f i∂gj

∂xi− gi

∂f j

∂xi

)∂

∂xj.

The definition of a Lie algebra over a field k is then the following:a Lie algebra is a vector space g over k equipped with a bilinear form[−,−] called the Lie bracket that is skew-symmetric and satisfies theJacobi identity:

[X, [Y,Z]] + [Z, [X,Y ]] + [Y, [Z,X]] = 0.

The Jacobi identity can be thought of as a "linearization of asso-ciativity".

We want to compute the commutator more explicitly. Let γX be acurve on G with γX(0) = e and γ′X(0) = X. Then for any g ∈ G, thecurve gγX satisfies (gγX)(0) = g and (gγX)′(0) = (delg)(X).

Hence we have

LX(f)(g) =d

dtf(gγX(t))

∣∣∣∣t=0

Then

(LXLY )(f)(g) =d

dtLY (f)(gγX(t))

∣∣∣∣t=0

=∂2

∂s∂tf(gγX(t)γY (s))

∣∣∣∣s=t=0

Hence

(def)([X,Y ]) = (LXLY − LY LX)(f)(e)

=∂2

∂s∂t(f(γX(t)γY (s))− f(γY (s)γX(t)))

∣∣∣∣s=t=0

(1)

5It is not immediately obvious that this is a derivation, but it is easily checked.

47

This is reasonable explicit, but can be made more so by workingwith coordinates. Choose local coordinates (x, U) around the identitye ∈ G. We can identity the codomain of x : U → Rn with g = TeG.We require that x(e) = 0 and dex = id (we can for example choosex = exp−1).

Now consider the map m : V × V → g given by (X,Y ) 7→x(x−1(X)x−1(Y )). This is well-defined for a small enough neighbour-hood U of 0 ∈ g.

Applying equation (??) to f = ` x (where ` is a linear functionalon g) we get

`([X,Y ]) =∂2

∂s∂t`(x(x−1(tX)(x−1(sY ))− x(x−1(sY )x−1(tX)))

∣∣∣∣t=s=0

= `

(∂2

∂s∂t(m(tX, sY )−m(sY, tX))

∣∣∣∣t=s=0

)Therefore

[X,Y ] =∂2

∂s∂t(m(tX, sY )−m(sY, tX))

∣∣∣∣t=s=0

This can further be written as follows: asm(0, 0) = 0, m(X, 0) = Xand m(0, Y ) = Y , the Taylor expansion of m at (0, 0) has the form

m(X,Y ) = X + Y +B(X,Y ) + h.o.t

where B : g× g→ g is a bilinear map [[WHY?????]]Hence in this description, we have that

[X,Y ] = B(X,Y )−B(Y,X).

Example 13.8. Let G = GL(V ) for V a finite-dimensional real orcomplex vector space. Choose coordinates by setting x(A) = A − I.Then

m(X,Y ) = (1 +X)(1 + Y )− 1 = XY +X + Y.

Hence in this case [X,Y ] = XY − Y X. F

The space EndV = Te GL(V ) with this bracket is denoted by gl(V ).With this done, here’s a theorem:

Theorem 13.9. If π : G → H is a Lie group homomorphism, thenπ∗ = deπ : g → h is a Lie algebra homomorphism. That is, the mapπ∗ preserves the bracket:

π∗([X,Y ]) = [π∗(X), π∗(Y )].

Also, if G is connected, π is completely determined by π∗.

48

Proof. Recall first that for all X ∈ g we have π(exp(tX)) =exp(tπ∗(X)). First off, we have, as above, that

LX(f)(g) =d

dt(f(g exp(tX)))

∣∣∣∣t=0

and one also verifies that

Lπ∗(X)(f)(π(g)) = LX(f π)(g).

This follows more or less directly from the definitions.[[[[ rest of the proof here ]]]]

49

14 Lie subgroupsBy definition, a closed Lie subgroup of a Lie group G is a subgroupH which is also a closed submanifold.

Then H is itself a Lie group. We can identify h = TeH with asubspace of g = TeG. As the embedding H → G is a Lie grouphomomorphism, this gives an identification of h with a Lie subalgebraof g.

In fact, the condition that H is a submanifold is redundant:

Theorem 14.1. If G is a Lie group and H is a closed subgroup, thenH is a closed submanifold.

Proof. This will be a bit tricky.Consider the subspace V ⊂ g consisting of X ∈ g such that there

exist Xn ∈ g and αn ∈ R such that exp(Xn) ∈ H and αnXn → X asn→∞ and also |αn| → ∞ as n→∞.

Note that V is non-empty since 0 ∈ V .Claim 1: If X ∈ V , then exp(tX) ∈ H for all t ∈ R.Take t ∈ R. As |αn| → ∞, we can find mn ∈ Z such that

mn

αn→ t

as t→∞ (take for example mn = [αnt]). Then

H 3 exp(Xn)mn = exp(mnXn)

= exp(mn

αnαnXn)

→ exp(tX).

This implies that exp(tX) ∈ H, since H is closed.Claim 2: The set V ⊂ g is a sub-vector space of g.By Claim 1, V is closed under multiplication by scalars. Therefore,

we have to show that if X,Y ∈ V , then X + Y ∈ V .Consider the map γ(t) = exp(tX) exp(tY ). Then γ′(0) = X + Y .

For small |t|, we can write γ(t) = exp(f(t)), where f is a smoothg-valued function, since exp is a local diffeomorphism at e.

Then f ′(0) = X + Y , so f(t)t → X + Y as t → 0. In particular,

nf( 1n )→ X+Y as n→∞. By definition of γ, we have exp(f( 1

n )) ∈ H.Now by definition of V , letting αn = n and Xn = f( 1

n ), we concludethat X + Y ∈ V .

Claim 3: H is a closed submanifold with TeH = V .Let V ′ be a complementary subspace to V in g, i.e. a subspace

such that g = V ⊕ V ′.Consider the map ϕ : V × V ′ → G defined by

ϕ(X,Y ) = exp(X) exp(Y ).

50

Then d(0,0)ϕ = idg, so we can find a neighbourhood Ω of 0 in V andΩ′ of 0 ∈ V ′ such that ϕ is a diffeomorphism of Ω×Ω′ onto ϕ(Ω×Ω′).

We will show that for n large enough, we have the equality

ϕ

(Ω× 1

nΩ′)∩H = exp(Ω× 0).

Suppose this is not true. Then we can find (Xn, X′n) ∈ Ω×Ω′ such

that exp(Xn) exp(X′nn ) ∈ H and X ′n 6= 0.

Take any limit point X ′ of the sequence X′n‖Xn‖ . Then X ′ 6= 0 and

X ′ ∈ V ′ and exp(X′nn ) ∈ H (since the left term is in H). Then also

n

‖X ′n‖· X′n

n→ X ′,

Then by definition of V , we must have X ′ ∈ V . This is a contradiction,since V ′ is closed.

Hence for some n, we have ϕ(Ω× 1

nΩ′)∩H = exp(Ω×0). Thus,

letting U = ϕ(Ω× 1

nΩ′), we can take as coordinate system x = ϕ−1,

such that

U ∩H = g ∈ H | the coordinates x(g) in V ′ are zero .

ThusH satisfies the definition of a closed submanifold near the identitye. For arbaitry h ∈ H, we can take the neigbourhood hU of h ∈ G andcoordinates xh(−) = x(h−1−).

Here’s a surprising consequence:

Theorem 14.2. If π : G → H is a continous homomorphism of Liegroup, then π is smooth.

In particular, for Lie groups, there is no difference between finite-dimensional representations of locally compact groups and Lie groups.

Proof. Consider the graph Γ of π:

γ = (g, π(g)) | g ∈ G ⊂ G×H.

Since π is continous, Γ is a closed subgroup of the product. Hence bythe theorem, Γ is a also a closed submanifold.

Consider the projection π : G×H → G, and let q = p1

∣∣Γ. Then q

is a smooth isomorphism between Γ and G.Since q is a Lie group homomorphism, it has constant rank (proof:

for all x ∈ Γ, we have q = lq(x)−1 q lx).Now, by the constant rank theorem, a constant rank bijection is a

diffeomorphism. Thus q−1 is smooth.But π is the composition of q−1 with p2 : G × H → H, so π is

smooth.

51

Proposition 14.3. If π : G→ H is a Lie group homomorphism, thenkerπ is a closed Lie subgroup of G with Lie algebra ker(π∗ : g→ h).

Proof. Clearly kerπ is closed, hence it is a closed subgroup. Hence it isa closed Lie subgroup by Theorem ??. So we need to show the secondstatement (let LieAlg(G) for the moment denote the Lie algebra of aLie group):

X ∈ LieAlg(kerπ)⇔ exp(tX) ∈ kerπ ∀t⇔ π(exp tX) = e∀t⇔ exp(tπ∗X) = e∀t⇔ π∗(X) = 0,

so X ∈ kerπ∗.

Proposition 14.4. Let π : G → GL(V ) be a representation of a Liegroup G on a finite-dimensional real or complex vector space V . Takeany v ∈ V . Then

H = g ∈ G |π(g)v = v

is a closed Lie subgroup of G with Lie algebra

h = X ∈ g |π∗(X)v = 0.

Proof. That H is closed Lie subgroup is immediate.Suppose X ∈ h. Then exp(tX) ∈ H for all t. But this means that

π(exp(tX))v = exp(tπ∗(X))v = v for all t. But differentiating bothsides give π∗(X)v = 0.

52

15 Classical Lie groupsThis section will give examples of Lie groups and their Lie alge-bras explicitly in terms of matrices. The classical Lie groups aregroups of linear transformations preserving some symmetric/skew-symmetric/Hermitian form on a real/complex/quaternionic space, aswell as their special versions.

15.1 The general linear groupLet V be a real or complex vector space of dimension n. Then GL(V )is the set of invertible linear operators V → V . If a basis for V isgiven, then GL(V ) is realized as the set of invertible n × n matriceswith determinant non-zero.

If V is a real vector space, then the dimension of GL(V ) is n2, andif V is complex, then the dimension is 2n2.

The Lie algebra of GL(V ), denoted by gl(V ) is identified withEnd(V ), or, if a basis is given, with Matn,n(V ), since GL(V ) is anopen subset in End(V ).

15.2 The special linear groupAgain, let V be a real or a complex vector space. The special lineargroup SL(V ) is by definition the subset of GL(V ) consisting of oper-ators of determinant 1. Since this is a closed condition, it follows fromTheorem ?? that SL(V ) is actually a closed Lie subgroup of GL(V ).Note that dim SL(V ) = n2 − 1 if V is a real vector space.

I claim that the Lie algebra of SL(V ) is the space of matrices inEnd(V ) of trace zero. To prove this, let At be an arc in SL(V )such that A0 = In and d

dt

∣∣t=0

At = X. Then Ate1 ∧ . . . ∧ Aten =e1 ∧ . . . ∧ en for all t since the determinant of At is 1 for all t. Thus,taking derivatives of both sides, we get

0 =d

dt

∣∣∣∣t=0

Ate1 ∧ . . . ∧Aten

=

n∑i=1

e1 ∧ . . . ∧Xei ∧ . . . en

= Tr(X)e1 ∧ . . . ∧ en.

In the second line we used the product rule for the derivative (whichworks just as well with exterior products (in fact, with any bilienaroperator). Hence Tr(X) = 0. Counting dimensions, we see that thespace of traceless matrices have the same dimension as SL(V ), so thatwe can identify sl(V ) with the space of traceless matrices.

53

15.3 Groups defined by preservation of a bilinearformLet V be as above and consider a bilinear B form on V . Let

G = g ∈ GL(V ) | B(gv, gw) = B(v, w) for all v, w ∈ V .

Then G is a subgroup of GL(V ). We shall verify that it is indeeda Lie subgroup.

LetW be the space of bilinear forms on V . Note thatW ' V ∗⊗V ∗.Define a representation π of GL(V ) on W by

(π(g)B′)(v, w) = B′(g−1v, g−1w).

Then G is exactly the set g ∈ GL(V ) | π(g)B = B, so that fromProposition ??, it follows that G is a Lie group. with Lie algebra

g = X ∈ gl(V ) | π∗(X)B = 0.

We want this a little more explicit. So note that we can write

π∗(X) =d

dtπ(exp(tX))

∣∣∣∣t=0

This is a simple calculation using the functoriality of πast and thedefinition of the exponential map. Hence we get

(π∗(X)B)(v, w) =d

dt(π(exp(tX))B)(v, w)

∣∣∣∣t=0

=d

dtB(e−tXv, e−tXw)

∣∣∣∣t=0

= −B(Xv,w)−B(v,Xw).

We used the chain rule in the last line.Thus

g = X ∈ gl(V ) | B(Xv,w) +B(v,Xw) = 0.

The dimension of G will of course depend upon the choice of B.

15.4 The orthogonal group O(n)

Let V = Rn and let B be the usual scalar product

(v, w) =

n∑i=1

viwi.

54

Then the group G for B as above is called the orthogonal group anddenoted by O(n), On(R) or O(n;R).

Note that we have

(Xv,w) + (v,Xw) = (Xv,w) + (XT v, w) = ((X +XT )v, w).

This inner product should be zero for all v, w. Hence we have:

o(n) = X ∈ gln(R) | X +XT = 0,

by the previous section. Thus the Lie algebra consists of all skew-symmetric matrices. We see that O(n) have dimension 1 + 2 + . . . +(n − 1) = n(n − 1)/2. The special orthogonal group SO(n) is thesubgroup of O(n) whose determinants are 1. In other words: SO(n) =O(n) ∩ SLn(R). This is the connected component of the identity ofOn(R). Then the Lie algebra is given by

sln(R) = X ∈ gln(R) | XT = −X, TrX = 0,

which is equal to o(n) since the condition XT = −X implies thatTrX = 0.

We can do the same for V = Cn with the same bilinear form. Thecorresponding groups are denoted by On(C) and SOn(C). These arehowever not compact anymore, since B ⊗ C is not positive definite.

15.5 Indefinite orthogonal groups O(k, l)

Let V = Rk+l for k, l ≥ 1 and consider the bilinear form

B(v, w) =

k∑i=1

viwi −k+l∑i=k+1

viwi

= (Av,w)

where

A =

Ik 0

0 −Il

.

The group of transformations preserving this form is then by definitionthe indefinite orhogonal group O(k, n). As above, we see that its Liealgebra is

o(k, l) = X ∈ glk+l(R) | AX +XTA = 0.

This can be made even more explicit. Let X ∈ o(k, l). Write

X =

P Q

R S

,

55

where P is a k × k matrix, Q is a l × k matrix, R is a k × l matrix,and S is a l × l matrix. Then one can check that the condition for Xto be in o(k, l) is equivalent to P = −PT , S = −ST and R = Q. Thusthe dimension of o(k, l) is

k(k − 1)

2+l(l − 1)

2+ kl =

n(n− 1)

2.

We also have its special version: SO(k, l) := O(k, l) ∩ SLk+l(R).

15.6 The unitary group U(n)

Let V = Cn, and consider the Hermitian form

〈v, w〉 =

n∑i=1

viwi.

The unitary group U(n) is the group of C-linear transformationspreserving this form. The proof in the case of O(n) goes through hereas well, so that we see that

u(n) = X ∈ gln(C) | X = −X∗,

where the star means the conjugate transpose of X. We also have thespecial unitary group SU(n) = U(n) ∩ SLn(C) with Lie algebra

su(n) = X ∈ gln(C) | X = −X∗, TrX = 0.

The dimension of U(n) is n2. To see this, note that matrices in theLie algebra consist of real entries along the diagonal, plus free entriesabove the diagonal (the entries below the diagonal are minus conjugatethose above). Thus the total freedom is n+ 2 · n(n−1)

2 = n2.

15.7 The symplectic Lie groups Sp(2n,R)Let V = R2n or V = C2n and consider

(v, w) =

2n∑i=1

(viwi+n − vi+nwi)

= (Jv,w)

where

J =

0 −In−In 0

.

56

The group of transformations preserving this form are the symplecticgroups Sp(2n,R) or Sp(2n,C). The Lie algebra is

sp(2n; k) = X ∈ gln(k) | JX +XTJ = 0.

In the same way is with the indefinite orthogonal group, one can com-pute the Lie algebra more explicitly. Namely, write X as

X =

A B

C D

.

Here A,B,C and D are n×n matrices. Here the condition on X to bein the Lie algebra is equivalent to D = −AT and B and C being skew-symmetric matrices. Hence the dimension is n2 + 2 · n

2−n2 = 2n2 − n.

15.8 The compact symplectic group Sp(n)

This is the group defined by Sp(n) = Sp(2n;C) ∩ U(2n) with Liealgebra

sp(n) = X ∈ gln(C) | X +X∗ = 0, JX +XTJ = 0,

where J is as above. It can be shown that Sp(n) is the subgroup ofGLn(H) preserving

∑ni=1 viwi.

57

16 Simply connected Lie groupsThere is a close correspondence between the category of Lie algebrasand the category of Lie groups. In particular, it is true that any Liesubalgebra correspond to a Lie subgroup with that Lie algebra.

To show this, we need to remind ourselves of some definitions frommanifold theory, however.

16.1 Some facts about distributionsLet M be a smooth manifold and let D ⊂ TM be a subbundle of thetangent bundle ofM . Then an immersed submanifold N ⊂M is calledan integral manifold if TpN = Dp for all p ∈ N .

Here’s the so-called Frobenius integrability condition:

Theorem 16.1. Suppose D is integrable. Then for all p0 ∈M , thereexists a coordinate chart (x, U) around p0 such that

Dp = Span

∂

∂x1

∣∣∣∣p

, . . . ,∂

∂xk

∣∣∣∣p

for all p ∈ U .

The proof is surprisingly short, and can be found in for exampleSpivak’s book [?].

This implies that there locally exists a unique integral manifoldpassing through a given point:

N = p ∈ U |xj(p) = xj(p0) for j = k + 1, . . . , n.

Uniqueness means that if N ′ is another integral manifold passingthrough p, then N ∩N ′ is open in N and N ′.

This implies that every point lies in a unique maximal connectedintegral manifoold, called a leaf of the foliation defined by D.

16.2 Back to Lie subgroupsTheorem 16.2. Let G be a Lie group with Lie algebra g. Let h ⊂ gbe a Lie subalgebra.

Then there exists a unique Lie subgroup H ⊂ G with Lie algebra h.

Proof. Consider the distribution D on G defined by Dg = (delg)(h).As h is a Lie subalgebra of g, it is integrable.

Since left translations map D into itself, they map leaves into leavesof the corresponding foliation.

It follows that if H is the leaf passing through e, then H = hHfor all h ∈ H. Therefore H ⊂ G is a subgroup. It is a Lie subgroupbecause it is locally closed.

58

Now, if K ⊂ G is another connected Lie subgroup with Lie algebrah, then K would be an integral manifold of D passing through e. ThusK is an open subgroup. But K is closed as well, since

K = H\

⋃h 6 inK

hK

.

As H is connected, H ⊂ K.

Proposition 16.3. Assume π : G→ H is a Lie group homomorphismwith G connected. Then π is completely determined by π∗ : g→ h.

Proof. Consider the subgroup K generated by exp g ⊂ G. Since exp gcontains an open neighbourhood of e, K is an open subgroup of G.But this implies that K is closed, hence K = G.

Thus, the statement is proved since π(expX) = expπ∗X, and everyelement of G is a product of terms looking like expX.

Remark. This proves that the functor LieAlg : LieGrps → LieAlgssending a Lie group to its Lie algebra is faithful.

It is not full, however. Take for example G = S1 and H = R.Then g = h = R, so HomLieAlg(g, h) = R, but HomLieGrp(S

1,R) = 06.

Recall that a map p : N →M is a covering map if p is continousand surjective and a local homeomorphism such that each point p ∈Mhas a neigbourhood U such that the pre-image decomposes as

p−1(U) =⊔i∈I

Ui

with p∣∣Ui

: Ui → U a homeomorphism.

Example 16.4. Suppose that π : G → H is a Lie group homomor-phism with H connected, such that π∗ : g → h is an isomorphism.Then π is a covering map:

We can find U ⊃ e ∈ G such that π∣∣U

: U → π(U) is a diffeomor-phism. Choose a neighbourhood V of e ∈ G such that V V −1 ⊂ U .

Note that as π is open and H is connected, π is surjective. Thenfor all g ∈ G, we have

π−1(π(g)π(V )) =⊔

h∈kerπ

hgV.

We have to check that h1g ∩ h2gV = ∅ for h1, h2 ∈ kerπ and h1 6= h2.6Proof: Any such homomorphism must have compact image. But the image must also

be discrete, since otherwise it would generate R. Hence the image is a finite subgroup ofR. But the only finite subgroup of R is 0.

59

Suppose h1gv1 = h2gv2 for some v1, v2 ∈ V . This is equivalent tog−1h−1

2 h1g = v2v−11 ∈ V V −1 ⊂ U . The left hand side is in the kernel

of π, but π is a homeomorhpism on U 3 e, hence g−1h−12 h1g = e,

hence h1 = h2. F

If a topological space is nice enough (for example a connected man-ifold), then there exists a unique universal covering space π : M →Msuch that if p : P → N is any covering space and f : M → N is acontinous map, then there exists a unique map f : M → P such thatp f = f π.

Furthermore, a covering map λ : Q→M is universal if and only ifQ is simply connected.

We can define M as the set of equivalence classes [γ] of curvesγ : [0, 1] → M such that γ1 ∼ γ2 if γ1(1) = γ2(1) and γ1 and γ2 arehomotopic within the class of curves ending at γ1(1) = γ2(1). This sethas a natural topology. See e.g. Munkres [?].

Proposition 16.5. Let G be a connected Lie group and π : G → Ga universal covering space. Then G has a natural structure of a Liegroup so that π becomes a Lie group homomorphism.

Proof. Suppose e ∈ π−1(e). Then using that π × π : G× G → G×Gis a universal cover, there exist a unique lift of the product map m :G×G→ G to a map m : G× G→ G such that m(e, e) = e.

Then G is a Lie group with inverse given by the lift of the inversesuch that e 7→ e.

Theorem 16.6. Let G,H be Lie groups with G simply connected.Then any Lie algebra homomorphism ρ : g → h integrates to a Liegroup homomorphism π : G→ H.

Proof. Consider the Lie subalgebra

p = (X, ρ(X)) |X ∈ g ⊂ g⊕ h.

Let P ⊂ G × H be the connected Lie subgroup with Lie algebra p.Consider the projection p1 : G × H → G and let q = p1

∣∣P. Then

q∗ : p→ g is an isomorphism.Hence q : P → G is a covering map. As G is already simplycon-

nected, q must be an isomorphism. Then we can define π as the compo-sition of q−1 : G→ P ⊂ G×H with the projection p2 : G×H → H.

Corollary 16.7. Two simply connected Lie groups are isomorphic ifand only if g ' h.

Here is a theorem we will only state, known as “Lie’s third fun-damental theorem” , which is due to Cartan.

60

Theorem 16.8. For any finite-dimensional real Lie algebra g, thereexists a simply connected Lie group G with Lie algebra g.

Thus the category of simply connected Lie groups is equivalent tothe category of Lie algebras.

One possible proof is based on Ado’s theorem, which we also willnot prove:

Theorem 16.9. Any finite-dimensional real Lie algebra is isomorphicto a Lie subalgebra of gln(R).

Given this, to prove Lie’s theorem, we can take g ⊂ gln(R). Thenlet G be the Lie subgroup of GLn(R) corresponding to g, and then passto the universal covering.

Remark. It is not true that every Lie group can be embedded inGLn(R). Here’s an example: let G be the universal cover of SL2(R).

One can show that SL2(R) is a fiber bundle with fiber R overR2\0, hence its fundamental group is Z.

Complexifying, we get the corresponding group SL2(C), which inthe same manner can be shown to have trivial fundamental group.

Now suppose we have a representation p : G → GLn(R) for somen. By inclusion, we have a representation G p→ GLn(R)→ GLn(C).

This is the same map as the composition

Gπ→ SL2(R)→ SL2(C)→ GLn(C)

where π is the covering space map. Since the map on tangent spaces isthe same. But π is not injective, hence p cannot be injective.

If G is connected but not simply-connected, then G ' G/Γ forsome discret normal subgroup Γ ⊂ G. It is easy to show that any suchsubgroup must be contained in the center of G.

Example 16.10. The groups SU(n), Sp(n) are simply connected.The group SO(n;R) is connected, but not simply connected. We

have π1(SO(2)) = π1(S1) = Z. For n > 2, we have π1(SO(n)) = Z/2.The universal covers are the “spin groups”.

We will prove (almost) this later. F

61

17 The adjoint representationLet G be a Lie group. For any g ∈ G consider the automorphismh 7→ ghg−1 of G. Thus we have a map G → Aut(G). Its differen-tial at the identity is denoted by Ad, and we get a homomorphismAd : G→ GL(g), called the adjoint representation.

The differential of Ad at e ∈ G gives a Lie algebra homomorphism

ad : g→ gl(g),

also called the adjoint representation.We can compute this map more explicitly. Let X,Y ∈ g. Then by

definition(adX)(Y ) =

d

dt(Ad exp(tX))(Y )

∣∣∣∣t=0

.

Let f ∈ C∞(G). Then

(def)(ad(X)(Y )) =d

dt(def)(Ad exp(tX)(Y ))

∣∣∣∣t=0

=∂2

∂t∂sf(exp(tX) exp(sY ) exp(−tX))

∣∣∣∣s=t=0

=∂2

∂s∂tf(exp(tX) exp(sY ))− f(exp(sY ) exp(tX))

∣∣∣∣s=t=0

= (def)([X,Y ]).

Therefore, (adX)(Y ) = [X,Y ].Of course the adjoint map ad : g→ gl(g) defined by X ·Y = [X,Y ]

is well-defined for any Lie algebra g. Note that the identity

ad[X,Y ] = [ad(X), ad(Y )] = ad(X) ad(Y )− ad(Y ) ad(X)

is exactly the Jacobi identity.The kernel of ad is exactly the center of g:

z(g) = X ∈ g | [X,Y ] = 0∀Y ∈ g.

Proposition 17.1. Let G be a Lie group and let X,Y ∈ g. Then thefollowing two conditions are equivalent:

1. [X,Y ] = 0.

2. exp(tX) exp(sY ) = exp(sY ) exp(tX) for all s, t ∈ R.Furthermore, if these conditions are satisfied, then exp(X + Y ) =

exp(X) exp(Y ).

Proof. This is Lemma 13 on page 157 in [?].

62

Proposition 17.2. Any connected abelian Lie group G is isomorphicto Rk × Tl for some natural numbers k, l ≥ 0.

Proof. Consider g as a Lie group under addition. Then exp : g→ G isa Lie group homomorphism by Proposition ??.

By Example ??, exp is a covering map. Let Γ be its kernel, whichmust a discrete subgroup of g.

But for any discrete subgroup Γ of g ' Rn, there is a basise1, . . . , ek such that Γ is equal to Ze1 ⊕ . . .⊕ Zek. Hence

g/Γ ' Rn/(Ze1 ⊕ . . .⊕ Zek) ' Tk × Rn−k.

63

18 Structure and representation theory ofSU(2)

Recall that the Lie algebra of SU(2) is

su(2) = X ∈ gl2(C) | X∗ +X = 0, Tr(X) = 0.

Recall also that

sl2(C) = X ∈ gl2(C) | Tr(X) = 0.

The latter is a complex vector space, and in fact we have su(2)⊗RC = sl2(C). Note that one inclusion is obvious. For the other, notethat any X ∈ sl2(C) can be written as

X =X −X∗

2+ i

X +X∗

2i,

namely as a sum of two matrices in su(2).There is a standard C-basis for sl2(C) given by the following three

matrices:

E =

0 1

0 0

F =

0 0

1 0

H =

1 0

0 −1

.

A basis for su(2) over R is E − F , i(E + F ) and iH. Note thatRiH ⊂ su(2) is the Lie algebra of the torus

T =

z 0

0 z

| z ∈ S1

.

The Lie algebra structure on sl2(C) is defined by the rules

[H,E] = 2E [H,F ] = −2F [E,F ] = H.

Lemma 18.1. The group SU(2) is simply-connected.

Proof. One checks that

SU(2) =

α β

−β α

| α, β ∈ C, |α|2 + |β|2 = 1

This is clearly homeomorphic to S3, which is simply-connected.

64

Therefore, from Theorem ??, we have a one to one correspondencebetween Lie group homomorphisms SU(2) → GL(V ) and Lie algebrahomomorphisms su(2)→ gl(V ).

Recall that if g is Lie algebra, then a representation of g on Vis the same thing as a Lie algebra homomorphism g → gl(V ). Wealso say that V is a g-module. We say that a subspace W ⊂ V isinvariant if X ·W ⊂ W for all X ∈ g. A representation g → gl(V )is irreducible (or the g-module V is simple) if there are no properinvariant subspaces.

It is an exercise to see that if G is connected Lie group, then thesenotions coincide with those of representation of G (as opposed to rep-resentations of g).

We now want to classify irreducible representations of SU(2) oncomplex finite-dimensional vector spaces V . Equivalently, we wantto classify irreducible representations of su(2) on complex finite-dimensional vectors spaces V .

Now, since su(2) ⊗R C ' sl2(C), any Lie algebra homomorphismsu(2) → gl(V ) (where we consider V as a real vector space) extendsuniquely to a Lie algebra homomorphism sl2(C) → gl(V ) of complexvector spaces.

So classifying su(2)-modules is the same as classifying sl2(C)-modules.

Assume now V is a finite-dimensional sl2(C)-module. Then thisexponentiates to a representation of SU(2):

su(2) //

sl2(C)

SU(2)

// SL2(C)

exp //

gl(V ) GL(V )

By restriction, the torus T ⊂ SU(2) act on V as well, and bycomplete reducibility, V decomposes as a sum of T -invariant subspaces.Since tiH | t ∈ R exponentiates to the torus T, V decomposes in asa sum of H-invariant subspaces as well.

Lemma 18.2. The eigenvalues of H are integers.

Proof. Note that each irreducible eigenspace must be one-dimensional,since on each eigenspace, H is sent to a diagonal matrix. Thus on eachone-dimensional irreducible representation, itH is sent to multiplica-tion by itα.

Also note that exp(itH) ∈ S1, so that the action of H on (C,+)lifts to a representation of S1. These are all of the form θ 7→ nθ.

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Exponentiating, we get that eitH = eitα for all t ∈ R. Put t = 2π.Then we get that 1 = e2πiα, which is possible only if α ∈ Z.

It is common to index the eigenvalues of H on V by half-integers.Thus for s ∈ 1

2Z, we put

V (s) = ξ ∈ V | Hξ = 2sξ.

Elements of V (s) are called vectors of weight s. Thus we write

V =⊕s∈ 1

2Z

V (s).

Now recall the commutation relations in sl2(C): [H,E] = 2E and[H,F ] = −2F .

Lemma 18.3. We have

EV (s) ⊂ V (s+ 1)

andFV (s) ⊂ V (s− 1).

In particular, if v 6= 0 is a vector of heighest weight, then Ev = 0.

Proof. Let v ∈ EV (s). Then v = Ew for w ∈ V (s). Then

Hv = HEw

= HEw − EHw + EHw

= [H,E]w + EHw

= 2Ew + 2sEw

= 2v + 2sv

= 2(v + 1)v.

So v ∈ V (s+ 1). The calculation with F is similar.

These two observations are enough to prove quite a lot about rep-resentations of sl2(C).

Lemma 18.4. Suppose V is a finite-dimensional sl2(C)-module andξ ∈ V (s)\0 satisfies Eξ = 0 (that is, ξ is of heighest weight). Then

1. s ≥ 0.

2. F 2s+1ξ = 0 and F kξ 6= 0 for k = 0, . . . , 2s.

3. EF kξ = (2s− k + 1)kF k−1ξ for all k ≥ 0.

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Proof. We start by proving number 3 by induction. The base casek = 0 gives Eξ = 0, which is clear by assumption.

Note that EF = [E,F ] + FE = H + FE. Note also that sinceξ ∈ V (s), we have F kξ ∈ V (s− k) by the previous lemma. Then:

EF k+1ξ = EFF kξ

= (H + FE)F kξ

= H(F kξ) + F (EF kξ)

= 2(s− k)F kξ + (2s− k + 1)kF kξ

= (2s− k + 2sk − k2)F kα.

Now note that this is what we want, because

(2s− (k + 1) + 1)(k + 1) = (2s− k)(k + 1) = 2sk + 2s− k2 − k.

Thus number 3) is proved.Now let n be the smallest number such that Fnξ = 0 (the smallest

weight). Then Fnξ = 0. Then it follows from 3) that

0 = EFnξ = (2s− n+ 1)nFn−1ξ.

Hence 2s− n+ 1 = 0, so n = 2s+ 1. This proves 2). This also proves1) because n ≥ 1.

In view of this lemma, for any s ∈ 12Z+ = 0, 1

2 , 1, . . ., define ansl2(C)-module Vs as follows: it is a vector space with basis elementsdenoted by

ξs−s, ξs−s+1, . . . , ξss,

and where H,F and E act as follows:

Hξst = 2tξst Fξst = ξst−1 Eξss−k = (2s− k + 1)kξss−k+1.

Then it is an (easy) exercise to see that this defines an sl2(C)-module. It is also easy to see that Vs is irreducible, since, startingfrom any basis vector ξst , we can use F and E to raise and lower theindices to generate all of Vs.

The number s in Vs is called the heighest weight or spin of Vs.The lemma tells us that if V is a finite-dimensional sl2(C)-module

and ξ ∈ V (s), ξ 6= 0 and Eξ = 0 (that is, ξ is a heighest weight vector),then we get an embedding Vs → V given by ξss−k 7→ F kξ.

Theorem 18.5. The sl2(C)-modules Vs (s ∈ 12Z) are irreducible, pair-

wise non-isomorphic, and any finite-dimensional irreducible sl2(C)-module is isomorphic to some Vs.

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Proof. The first statement is clear, since all the Vs have different di-mension. Now, if V is an irreducible sl2(C)-module, the observationabove tells that us we have an embedding Vs → V .

Since sl2(C) is semisimple, it follows from Weyls theorem ([[ref??]]),that every invariant subspace has a complementary invariant subspace.But V was irreducible, so Vs = V .

Example 18.6. Let s = 0. Then dimV0 = 1, and this corresponds tothe trivial representation of SU(2). F

Example 18.7. Let s = 12 . Then dimV 1

2= 2, and one sees that

H,F and E act precisely as the canonicial representation of sl2(C) onC2. F

Example 18.8. Let s = 1. Then dimV1 = 3, and V1 is the adjointrepresentation of sl2(C) with heigest weight vector ξ1

1 = E. F

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19 ToriA torus is a Lie group isomorphic to Tn = (S1)

n.Recall that by Proposition ??, any compact connected abelian Lie

group is a torus.

Proposition 19.1. Let G be a compact connected Lie group. Thenthere exists a maximal torus in G. Furthermore, we have a one to onecorrespondence between the maximal tori in G and the maximal abelianLie subalgebras of g.

Proof. As g is finite-dimensional, maximal abelian subalgebras exist.Let a be such a maximal abelian subalgebra of g. Then exp a is aconnected abelian Lie subgroup of G.

ThenH = exp a is a closed connected abelian subgroup of G. HenceH is a torus. Then h is abelian and h ⊃ a, hence h = a. HenceH = exp a, since the inclusion a ⊂ H must be a covering map.

On the other hand, if T ⊂ G is a maximal torus, then t ⊂ g isan abelian Lie algebra, hence it is contained in a maximal abelian Liealgebra a. Then exp a is a torus containing T , hence T = exp a, andt = a.

Our goal in this section is to show that any two maximal tori areconjugate. This is not so difficult in concrete cases:

Example 19.2. Let G = U(n). Then

T =

z

. . .

z

| z ∈ C

∩ SU(n) ' Tn

is a torus.Suppose H ⊂ G is a torus. As H is compact abelian, Cn decom-

poses into a direct sum of 1-dimensional H-invariant subspaces. Hencethere exist g ∈ U(n) such that gHg−1 ⊂ T , hence there exist g ∈ U(n)such that gHg−1 ⊂ T .

This shows that T is maximal and any maximal torus is conjugateto T . F

We need some preparations to prove the general case.If T is a torus, we say that g ∈ T is a topological generator if

gn | n ∈ Z = T.

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Theorem 19.3 (Kronecker). Consider T = Tn. Then

g =(e2πit1 , . . . , e2πitn

)is a topological generator if and only if the numbers 1, t1, . . . , tn arelinearly independent over Q.

Proof. Consider H = gn | n ∈ Z. If H 6= T , then T/H is a non-trivial compact abelian group, hence T/H has a nontrivial character.By composition, T has a non-trivial character χ such that χ

∣∣H

= 1(constantly the identity).

Note that G/H is trivial if and only if for every character χ, theimplication χ(g) = 1⇒ χ = 1 holds.

We have, for any character,

χ(e2πit1 , . . . , e2πitn

)= e2πim1t1 · . . . · e2πimntn

for some m1, . . . ,mn ∈ Z. So suppose χ(g) = 1. Then the mi’sconstitute an equation of linear dependence among the ti. But if theseare linearly independent, then χ is trivial.

Corollary 19.4. For any torus T , the set of topological generators isdense in T .

Proposition 19.5. Let G be a compact connected Lie group. LetT ⊂ G be a maximal torus. Then

1. N(T )∆= g ∈ G | gTg−1 = T is a closed subgroup of G.

2. N(T ) = T .

3. The group W (G,T )∆= N(T )/T is finite. It is called the Weyl

group of G.

Proof. It is obvious that N(T ) is a closed subgroup, since the definitionis a closed condition.

For 2., consider the homomorphism

α : N(T )→ Aut(T )

g 7→ α(g)(h) = ghg−1.

The automorphism group of the torus T is GLn(Z) ⊂ GLn(R).The group GLn(Z) is discrete, so α is locally constant. Thus

N(T ) ⊂ kerα. Thus gh = hg for all g ∈ N(T ) and all h ∈ T .Let X be an element of the Lie algebra of N(T ). Then exp(tX) com-mutes with T for all t ∈ R. It follows that the closure of the groupgenerated by exp(tX) is a compact connected abelian Lie group, hencemust be a torus.

70

But since T is maximal, this torus coincides with T , so exp(tX) ∈ Tfor all t ∈ R.

Thus the Lie algebra of N(T ) coincides with the Lie algebra of T .Hence N(T ) = T .

For 3., note that by 2., we have

N(T )/T = N(T )/N(T ),

which is finite because N(T ) is compact.

Example 19.6. Let G = U(n). And let T consist of the unitarydiagonal matrices. It is a maximal torus.

Then N(T ) consist of all the matrices g ∈ U(n) such that everyrow and column contains exactly one nonzero coefficient.

Thus N(T )/T ' Sn. F

19.1 Digression on Riemannian geometryRecall that a Riemannian manifold is a manifold M such that TpMis equipped with a scalar product depending smoothly on p. Thus, ifX,Y ∈ X(M) are two vector fields, then the map p 7→ (Xp, Yp) is asmooth map. Equivalently, a scalar product defines a smooth sectionof S2(T ∗M).

Given a piecewise C1-curve, we define its length by

L(γ)∆=

∫ b

a

‖γ′(t)‖dt.

Then we can define a metric on M (assuming M is connected), by

d(a, b) = infL(γ) | γ : [0, 1]→M piecewise C1 , γ(0) = a, γ(1) = b.

It is not difficult to see that the topology on M coincides with thetopology defined by this metric d.

A map γ : [a, b]→M is called a geodesic if for any t0 ∈ (a, b), thereexists a δ > 0 such that d(γ(s), γ(t)) = |s− t| for s, t ∈ (t0 − δ, t0 + δ).

We also allow linear change of variables, so t 7→ γ(at) is also calleda geodesic for α > 0.

Here is a theorem we wont prove.

Theorem 19.7. Let M be a Riemannian manifold. Then:

1. Any geodesic is smooth.2. For any p ∈M and X ∈ TpM , X 6= 0, there is a unique maximal

geodesic γ such that γ(0) = e and γ′(0) = X.3. (Hopf-Rinow theorem) If M is compact and connected, then any

maximal geodesic is defined on the whole line R, and ofr anyp, q ∈M , there exist a geodesic passing through p and q.

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19.2 Returning to toriFrom now on, assume G is a compact connected Lie group. Considerthe adjoint representation Ad: G→ GL(g). Since g is compact, thereexists an AdG-invariant scalar product (see Lemma ??). Then Ad :G → O(g, 〈−,−〉). Recall that the Lie algebra of O(g) consist of T ∈gl(g) such that

〈Tx, y〉+ 〈X,Ty〉 = 0

for all x, y ∈ g.Now we use left- or right-translations to define a scalar product on

TgG for any g ∈ G. Thus G becomes a Riemannian manifold such thatboth left and right translations preserve the Riemannian structure.

Theorem 19.8. The geodesics in G are translations (left or right) ofone-parameter subgroups of G.

Proof. The result is true for tori, since they are of the form Rn/Zn, sotheir geodesics are lines.

In general, it suffices to consider geodesics passing through e. Letγ : R → G be a geodesic with γ(0) = e. Put X = γ′(0) ∈ g. For anyg ∈ G, the map t 7→ gγ(t)g−1 is also a geodesic (because the scalarproduct is invariant under left- and right-translations). Its derivativeat 0 is Ad(g)(X).

Let T be a maximal torus containing exp(tX) for t ∈ R. ThenAd(g)(X) = X for all g ∈ T .

It follows that γ(t) = gγ(t)g−1 for all t ∈ R, since by point 2. inTheorem ??, geodesics are unique. In particular, γ(t) ∈ N(T ) = T .

Thus Imγ ⊂ T . Hence γ is a geodesic in T , where the result is true.Hence γ(t) = exp(tX), since the theorem is true for T .

The next result is completely fundamental for everything that fol-lows.

Theorem 19.9 (Cartan). Let G be a compact connected Lie groupand T ⊂ G a maximal torus. Then for all g ∈ G there exists hg ∈ Gsuch that hggh−1

g ∈ T .

Proof. Let t0 ∈ T be a topological generator of T . Let H0 ∈ t be suchthat exp(H0) = t0.

Choose a geodesic γ passing through e and g with γ(0) = e. LetX = γ′(0). Then γ(t) = exp(tX) by the previous theorem.

Consider the function

G 3 h 7→ 〈Ad(h)(X), H0〉 ∈ R

This is a function from a compact space to the real line, so it hasa maximum, achieved at, say, h ∈ G.

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Claim: [(Adh)(X), H0] = 0.Assuming the claim, it follows that exp(t(Adh)(X)) =

h exp(tX)h−1 commutes with exp(H0) = t0 for t ∈ R. As t0 is a topo-logical generator of T , it follows that h exp(tX)h−1 commutes withT .

As T is a maximal torus, it follows that h exp(tX)h−1 ∈ T for allt ∈ R. In particular, hgh−1 ∈ T .

Thus it remains to prove the claim. Take Y ∈ g. Then the function

t 7→ 〈Ad(exp(tY )h)(X), H0〉

attains its maximum at t = 0. Therefore its derivative at 0 is zero. So

0 = 〈ad(Y )(Adh)(X), H0〉= −〈[Ad(h)X,Y ], H0〉= 〈Y, [Ad(h)(X), H0]〉.

This is true for all Y ∈ g, hence [Ad(h)(X), H0] = 0.

Corollary 19.10. Any two maximal tori in G are conjugate.

Proof. Let T1 and T2 be two maximal tori. Let t1 ∈ T1 be a topologicalgenerator. Then there exists h ∈ G such that ht1h−1 ∈ T2. But thenhT1h

−1 ⊂ T2, since t1 was a topological generator. So T1 ⊂ h−1T2h.But T1 was maximal, so T1 = T2.

Corollary 19.11. Any element of G is contained in a maximal torus.In particular, exp: g→ G is surjective.

Proof. Let g ∈ G, and let T be a maximal torus. Then hgh−1 ∈ T forsome h ∈ G. But then g ∈ h−1Th, which is maximal.

To see the "in particular", let g ∈ G. Then g is contained in amaximal torus. Then the geodesic starting at e and ending at g musthave the form exp(tX).

Corollary 19.12. Any maximal torus T in G is a maximal abeliansubgroup of G.

Proof. Assume g commutes with T . Consider

H = C(g) = h ∈ G | hg = gh,

the centralizer of g in G. This is a compact connected Lie group. SinceT is connected, T ⊂ H. As g is contained in a maximal torus in G, wehave g ∈ H.

As T is a maximal torus in G, it is a maximal torus in H. Hencethere exists h ∈ H such that hgh−1 ∈ T . But g ∈ Z(H), henceg ∈ T .

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Remark. This implies that the map

α : W (G,T )→ Aut(T )

g 7→ α(gT )(h) = ghg−1

has trivial kernel. Hence W (G,T ) → Aut(T ).

74

20 RootsFirst we recall some notation from the representation theory of SU(2).There we had a maximal torus, called T , which was given by

T =

z 0

0 z

| z ∈ S1 ⊂ C

.

The Lie algebra of SU(2) is contained in the Lie algebra of SL(2,C).We write su(2) ⊂ sl2(C). The Lie algebra is spanned by [[write basishere]]

20.1 RootsLet G be a compact connected Lie group. Fix a maximal torus T .Consider gC = C⊗Rg, the complexification of the Lie algebra ofG. Thisis a complex Lie algebra, where we define [c⊗X, d⊗ Y ] = cd⊗ [X,Y ].

We can identifty g with the elements of the form 1⊗X for X ∈ g.Sometimes we write elements of gC as X + iY .

The adjoint representation Ad : G→ GL(g) extends to a represen-tation G→ GL(gC), which we will also denote by Ad.

Consider Ad∣∣T. As T is a compact abelian group, we have a decom-

position of Ad∣∣Tinto isotypic components corresponding to characters

χ ∈ T . Thus we can write

gC =⊕χ∈T

gC(χ)

where

gC(χ) = X ∈ gC | (Ad g)X = χ(g)X for all g ∈ T

is the eigenspace of X.We view T = S1 as a subgroup of C∗ = GL1(C). In this picture,

the Lie algebra of T is iR ⊂ C.Let t be the Lie algebra of T . Any character χ ∈ T is determined

by its differential χ∗ : t → iR, so we can think of χ∗ ∈ it∗, wheret∗ = HomR(t,R).

Then the set of differentials χ∗ (χ ∈ T ) form a subset X∗(T ) ⊂ it∗.It is a lattice in it∗ (here by lattice we mean a discrete subgroup ofmaximal rank), called the character lattice.

Indeed, if T = Rn/Zn, then t = Rn, so t∗ = Rn, and an element(it1, . . . , itn) ∈ it∗ defines a character Rn → T by

(a1, . . . , an) 7→ eia1t1 · . . . · eiantn ,

75

which factors through Rn/Zn if and only if all the ai ∈ 2πZ. ThenX∗(T ) = 2πiZn ⊂ iRn.

Consider now the complexification tC of the Lie algebra of T . Wedenote this complexification by h. It is called a Cartan subalgebraof gC. It is a “maximal toral subalgebra”.

Functionals t → R can be extended by complex linearity to h, soit∗ ⊂ h∗ = HomC(h,C). We then have

gC(χ) = X ∈ gC | (Ad g)X = χ(g), X ∀g ∈ T= X ∈ gC | [Y,X] = χ∗(Y )X, ∀Y ∈ T= X ∈ gC | [Y,X] = χ∗(Y )X, ∀Y ∈ h.

For α ∈ h∗, denote

gα = X ∈ gC | [H,X] = α(H)X∀H ∈ h.

In our earlier notation, this says that gα = gC(χ), where we think ofχ ∈ it∗ ⊂ h∗.

Lemma 20.1. We have that g0 = h.

Proof. This follows because t is a maximal abelian Lie subalgebra ofg, so that

t = X ∈ g | [H,X] = 0 ∀H ∈ t.

Now complexify both sides.

The elements α ∈ X∗(T )\0 such that gα 6= 0 are called the rootsof G or gC. We denote the set of roots by ∆.

To remember the position of g inside gC, it is convenient to intro-duce an involution on gC, defined by

(X + iY )? = −X + iY,

so that g = X ∈ gC | X = −X?.7This induces an involution α 7→ α in h∗ as well. Namely, define

α(H)∆= α(H?).

This is actually an involution, because

α(H) = α(H?) = α(H??) = α(H).

In fact, this definition ensures that α is complex linear as well: we havethat α(zH) = zα(H) (this is an easy but somewhat tedious exercise).

7Note that I have chosen to write X? for the involution to distinguish it from takingduals.

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This lets us recognice it∗ inside h∗ as well. Namely,

it∗ = α ∈ h∗ | α(t) ⊂ iR= α ∈ h∗ | α = α,

as those functionals fixed by the involution.

20.2 Some lemmasThe rest of this lecture will consist of several small results.

Lemma 20.2. We have a decomposition

gC = h⊕⊕α∈∆

gα.

Also:1. [gα, gβ ] ⊂ gα+β.2. g?α = g−α. Here g?α = X? | X ∈ gα.

Proof. The decomposition of gC is just that of writing gC into simul-taneous eigenspaces for the action of h, where h has zero eigenvalues,as proved above.

We have that

gα = X ∈ gC | [H,X] = α(H)X for all H ∈ h.

Let Y ∈ gβ . Then we look at [H, [X,Y ]]. By the Jacobi identity, thisis equal to

[H, [X,Y ]] = −[Y, [H,X]]− [X, [Y,H]]

= −α(H)[Y,X] + [X, [H,Y ]]

= α(H)[X,Y ] + β(H)[X,Y ]

= (α(H) + β(H))[X,Y ].

Hence [X,Y ] ∈ gα+β .For ii), note that α(t) ⊂ iR. Hence for H ∈ t, we have

α(H) = α(H?) = −α(H) = α(H).

Hence we have

gα = X ∈ gC | [H,X] = −α(H)X ∀H ∈ h= X ∈ g | [H,X] = −α(H)X ∀H ∈ it∗ ⊗R C= X ∈ g | [H,X?] = α(H)X? ∀H ∈ it∗ ⊗R C.

In the last equality, we used the involution on both sides to get

[H?, X?] = −[H,X] = α(H)X = α(H)X?,

since X is real (which gives X? = −X) and H? = H.

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20.3 An Ad-invariant inner productTo proceed, it is convenient to introduce an Ad-invariant scalar producton g. These exist:

Lemma 20.3. Let G be compact Lie group. Then there is an Ad-invariant inner product (−,−) on g, in the sense that

(g ·X, g · Y ) = (X,Y )

for X,Y ∈ g, where g ·X denotes the adjoint action Ad : G→ GL(g).This condition further implies that

([X,Y ], Z) + (Y, [X,Z]) = 0.

Proof. Choose any scalar product 〈−,−〉 on g and a normalized Haarmeasure dg on G. Then

(X,Y ) =

∫G

(g ·X, g · Y )dg

is an Ad-invariant scalar product on g. To see the other equality, letγ : R→ G be an arc with γ(0) = e ∈ G and γ′(0) = Z ∈ g. Then

0 =d

dt(γ(t) ·X, γ(t) · Y )

∣∣∣∣t=0

= limt→0

1

t((γ(t) ·X, γ(t) · Y )− (X, γ(t) · Y ) + (X, γ(t) · Y )− (X,Y ))

= limt→0

(1

t(γ(t) ·X −X), Y

)+

(X,

1

t(γ(t) · Y − Y )

)= ([Z,X], Y ) + (X, [Z, Y ])

Renaming X,Y, Z appropriately gives the required identity.

It will be convenient to require that (−,−) is negative definite on g.We can extend this form to gC be complex linearity. It will no longerbe definite (it doesn’t make sense), but it will still be non-degenerate(easy exercise).

Lemma 20.4. We have(gα, gβ) = 0

if α 6= −β.

Proof. This follows from the invariance of Ad. Namely let Xα ∈ gαand Xβ ∈ gβ . Then:

0 = ([H,Xα], Xβ) + (Xα, [H,Xβ ])

= (α(H)Xα, Xβ) + (Xα, β(H)Xβ)

= (α(H) + β(H))(Xα, Xβ)

for all H ∈ h.

78

By duality we can define a symmetric bilinear form on h∗ as well:let α ∈ h∗. Define its dual element hα by the equality

(hα, H) = α(H)

for all H ∈ h. To see that this is well-defined, choose an orthogonalbasis ei for h and let e∗i be the dual basis (with respect to the form(−,−)). Then the vector hα is just the transpose of the vector α.

We define(α, β) = (hα, hβ).

Lemma 20.5. We haveh?α = hα

for every α ∈ h∗.

Proof. First off, note that by definition (X?, Y ) = (X,Y ?). Then

(h?α, H) = (hα, H?) = α(H?) = α(H) = α(H) = (hα, H).

Here we used the definition of α as α(H) = α(H?).

It follows that if for α ∈ h satisfies α = α, then h?α = hα, whichimplies that hα ∈ it. Since (−,−) was negative definite on g, it followsthat (−,−) is positive definite on it.

Now let α ∈ ∆ be a root. Then define

Hα =2hα

(α, α)∈ it ⊂ h.

Lemma 20.6. Let α ∈ ∆, and X ∈ gα. Then

[X,X?] = (X,X?)hα.

Proof. We know already from Lemma (??) that [X,X?] ∈ h. Letβ ∈ h?. Then

β([X,X?]) = (hβ , [X,X?])

= −(hβ , [X?, X])

= ([X?, hβ ], X)

= (α(hβ)X?, X) = (X?, X)β(hα).

This holds for all β ∈ h∗, which implies the statement.

Choose a non-zero Eα ∈ gα, and normalize it so that (Eα, E?α) =

2(α,α) . Then

[Eα, E?α] = Hα.

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Put Fα = E?α. We thus get

[Hα, Eα] = 2Eα [Hα, Fα] = −2Fα [Eα, Fα] = Hα.

This defines a sub-Lie-algebra isomorphic to sl2(C). Thus for everyα ∈ ∆ we get an inclusion of Lie algebras

iα : sl2(C)→ gC.

Lemma 20.7. a) The spaces gα are one-dimensional.

b) Suppose α ∈ ∆ and c ∈ R? and cα ∈ ∆. Then c = ±1.

Proof. We can suppose α to be minimal, in the sense that cα 6∈ ∆ ifc ∈ (0, 1). Now consider the space

V = CFα ⊕ h⊕

⊕c>0,cα∈∆

gcα

.

This is a finite-dimensional sl2(C)-module. But we know from [[previ-ous lecture]] that their spectra are symmetric about the origin, countedwith multiplicities. Hence the big term must be isomorphic with theleft term, which is one-dimensional, and we must have c = 1.

Remark. The isomorphism h∗ ' h depends upon the choice of in-variant form, but the elements Hα (α ∈ ∆) does not depend uponthis choice: Hα is the unique element of the one-dimensional space[gα, g−α] ⊆ h such that α(Hα) = 2. Then Eα ∈ gα is defined up toscalar of modulus 1, such that [Eα, E

∗α] = Hα.

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21 Root systemsIn this lecture, we study root systems in the abstract.

Let V be a finite-dimensional Euclidean vector space (that is, areal vector space equipped with a scalar product). A finite subset∆ ⊂ V \0 is a root system if:

1. For every α ∈ ∆, the reflection sα with respect to the hyperplaneα⊥ leaves ∆ invariant.Note that

sα(v) = v − 2(α, v)

(α, α)α.

2. The number2(α, β)

(α, α)

is an integer for all α, β ∈ Z.

A root system is called reduced, if in addition the following conditionis satisfied:

3. If α, cα ∈ ∆, where c ∈ R×, then we must have c = ±1.

Remark. Number 3. is often assumed to be part of the definition of aroot system. It is also often assumed that ∆ span the whole space V ,but for us it will be important not to assume this.

Example 21.1. Here’s an example of a root system. It is denoted byA2 (and is associated with SU(3)). It is a regular hexagon.

Figure 1: The root system A2.

To check the conditions, we can choose the vertices αi to be givenby αi = (cos θi, sin θi) with θi = π

3 i for i = 0, . . . , 5π/3. Then we havethat

(cos θi, sin θi) · (cos θj , sin θj) = cos(θi − θj) = ±1

2or ± 1,

since θi − θj is a multiple of π/3. In either case, the second conditionis fulfilled. F

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We say that two root systems (V1,∆1) and (V2,∆2) are isomor-phic if there exists a linear isomorphism T : V1 → V2 such thatT (∆1) = ∆2 and Tsα = sT (α)T for all αin∆1. Note that we donot require the isomorphism to preserve the scalar product.

Theorem 21.2. Let G be a compact connected Lie group and T ⊂G a maximal torus. Then (it∗,∆) is a reduced root system. Up toisomorphism, this root system does not depend upon the choice of Tand the choice of an invariant form on g.

Proof. LONG PROOF HERE

21.1 Simple roots, Weyl chambers, and the WeylgroupLet (V,∆) be a root system and assume that Span∆ = V . We saythat a subset Π ⊂ ∆ is a system of simple roots if (1) Π is a basisof V and (2) that if β =

∑α∈Π cαα ∈ ∆, then either all cα ≥ 0 or all

cα ≤ 0.Reflections sα defined by simple roots are called simple reflec-

tions.If we fix a set of simple roots Π, then the roots β =

∑α cαα ∈ ∆

with positive cα are called positive roots. We denote the set ofpositive roots by ∆+. Thus ∆ = ∆+ ∪ −∆−.

It is a fact that every root system possesses systems of simple roots.In order to construct Π, we introduce the notion of a Weyl chamber:consider the set

V \⋃α∈∆

α⊥.

The connected components of this set are called the (open)Weyl cham-bers of ∆. Let C be such a Weyl chamber.

By definition, (α, β) 6= 0 for all α ∈ ∆ and β ∈ C. Hence for anyα ∈ ∆ we have either (α, β) > 0 for all β ∈ C or (α, β) < 0 for allβ ∈ C. In the first case, we say that α is C-positive. Let Π(C) ⊂ ∆be the set of C-positive roots α such that it is impossible to write α asβ + γ for C-positive roots β, γ ∈ ∆.

Proposition 21.3. The set Π(C) is a system of simple roots. Further-more, the assignment C 7→ Π(C) is a bijection between Weyl chambersand systems of simple roots.

Proof. By definition of Π(C), any C-positive root β has the form∑α∈Π(C) cαα for cα ∈ Z+ (why: because Π(C) is exactly the set of

C-positives that are such sums with only one term).It follows that Span Π(C) = V , since any cone of the same dimen-

sion as V span V .

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Therefore we have to show that Π(C) is a basis. Observe first thatif α, β ∈ Π(C), then α − β 6∈ ∆. Indeed, otherwise, either α − β ogβ−α is C-positive. But then α = β+ (α−β) or β = α+ (β−α) givesa contradiction to the definition of Π(C).

Observe next that if α, β ∈ Π(C), α 6= β, then (α, β) ≤ 0. Indeed,assume that (α, β) > 0. We may assume that ‖α‖ ≥ ‖β‖. Then

2(α, β)

(α, α)<

2‖α‖‖β‖‖α‖2

=2‖β‖‖α‖

≤ 2.

But the left term is a positive integer, hence it must be equal to 1. Butthen sα(β) = β − α ∈ ∆, contradicting the previous observation.

Assume now that∑α∈Π(C) cαα = 0 for some cα ∈ R. Consider the

sets

A = α ∈ Π(C) | cα ≥ 0, B = Π(C)\A,

and let v =∑α∈A cαα =

∑α∈B(−cα)α. Then

(v, v) =∑

α∈A,β∈B

cα(−cβ) · (α, β) ≤ 0,

Since the coefficients are positive by definition of A and B, and (α, β) ≤0 by the above observation. Hence v = 0.

Now take any β ∈ C. Then

0 = (v, β) =∑α∈A

cα(α, β) =∑α∈B

(−cα)(α, β)

The coefficients in the last equality are strictly negative and the dotproducts are strictly positive (since the αs are C-positive). Hence wemust conclude that B = ∅. Likewise we conclude that cα = 0 for allα ∈ A. Thus Π(C) is a basis, and Π(C) is a set of simple roots.

Now we have to show that the assignment is a bijection.Note that for any Weyl chamber C, we have

C = β ∈ V | (α, β) > 0 for any C-positive root α= β ∈ V | (α, β) > 0 ∀ α ∈ Π(C).

Thus we see that we can recover C from Π(C), so C 7→ Π(C) is injec-tive.

Assume now that Π is a system of simple roots. As Π is a basis,there exists a β0 ∈ V such that (α, β0) = 1 for all α ∈ Π. Then(α, β0) 6= 0 for all α ∈ ∆, hence β0 lies in a uniquely defined Weylchamber C.

Then we claim that Π ⊂ Π(C). Indeed, if α ∈ Π is equal toα = β + γ for some C-positive β, γ, then β and γ are positive withrespect to Π, as (β, β0) > 0 and (γ, β0) > 0.

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As β, γ cannot be proportional to α (∆ is reduced), decomposingβ and γ as

β =∑δ∈Π

c′δδ

andγ =

∑δ∈Π

c′′δ δ

with c′δ, c′′δ ≥ 0, we have

α =∑δ∈Π

(c′δ + c′′δ )δ,

contradicting the assumption that Π is a basis. But #Π = #Π(C), sothey must be equal.

Given a Weyl chamber C, we say that a hyperplane L ⊂ V is a wallof C if L ∩ C has nonempty interior in L (in the subspace topology).

Here are som facts that were given as an exercise:

• For all α ∈ ∆, the hyperplane α⊥ is a wall of some Weyl chamber.

• For any Weyl chamber C, the walls of C are the hyperplanes α⊥with α ∈ Π(C).

• This implies that any α ∈ ∆ lies in Π(C) for some Weyl chamberC.

By definition, the Weyl group W of (V,∆) is the subgroup ofO(V ) generated by the reflections sα for α ∈ ∆.

Since the reflections preserve the root system (which is a finite set),the Weyl group is itself finite. Note also that since the Weyl group acton the set of hyperplanes α⊥ for α ∈ ∆, it also act on the set C ofWeyl chambers.

Theorem 21.4. We have:

1. W acts freely and transitively on the set of Weyl chambers.

2. For any Weil chamber C, W is generated by the reflections sαfor α ∈ Π(C).

Half-proof: Let C be a Weyl chamber and let C ′ be an adjacent Weylchamber, meaning that there exists a hyperplane L such that L∩C∩C ′has nonempty interior in L.

Then L = α⊥ for some α ∈ Π(C) and C ′ = sα(C). Thensα(Π(C)) = Π(C ′).

It follows that for any β′ = sα(β) ∈ Π(C ′) (with β ∈ Π(C)), wehave

sβ′ = ssα(β) = sαsβsα,

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since for any orthogonal transformation we have sTα = TsαT−1.

Thus reflections sβ′ for β′ ∈ Π(C ′) lie in the group generated bysα, α ∈ Π(C).

Now, for any Weyl chamber C ′′ we can find a sequence of adjacentWeyl chambers C = C0, . . . , Cn = C ′′. Using the above argumentand induction on n, we conclude that there exists w ∈ W such thatw(C) = C ′′ and the reflections sβ′′ for β′′ ∈ Π(C ′′) lie in the groupgenerated by sα for α ∈ Π(C).

As any root in ∆ lies in some Π(C ′′), we see that W is generatedby sα, α ∈ ∆.

This proves that the action of W on the set of Weyl chambers istransitive. We will not prove nor need the fact that the action is free.For root spaces coming from compact groups, this will follow from theproof of the next result.

Example 21.5. Again consider the root system A2.One computes that the Weyl chambers are the interiorsof the rays generated by

(cos(π6 + π

3 k), sin(π6 + π3 k))

and(cos(π6 + π

3 (k + 1)), sin(π6 + π3 (k + 1))

)for k = 0, . . . , 5.

Let C be the Weyl chamber generated by the first two rays. Thenone computes that there are three C-positive roots: α1, α2, α3 (thefirst one of angle zero, next of angle 60, and last of angle 120). Butα2 = α1 + α3, so we discard it. Thus the C-simple roots of A2 are α1

and α2.We see that the Weyl group is the group D6 of symmetries preserv-

ing the hexagon. F

21.2 Root systems of compact Lie groupsAssume now that G is a compact connected Lie group and T ⊂ G amaximal torus. Consider the corresponding root system ∆ on V =it∗ ⊂ h∗.

Let V0 = SpanR ∆. Then for any α ∈ ∆, the reflection sα acttrivially on V ⊥0 . It follows that the group generated by sα (α ∈ ∆) canbe identified with the Weyl group W generated by sα

∣∣V0.

We have a homomorphism

r : N(T )/T → O(V )

defined by r(gT )(α)α (Ad g−1).

Theorem 21.6. The homomorphism r defines an isomorphism

N(T )/T 'W,

so that the two definitions of the Weyl group are equivalent.

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22 Cartan matrices and Dynkin diagramsLet (V,∆) be a reduced root system with Span ∆ = V . Fix a systemΠ of simple roots, say Π = α1, . . . , αr. The r × r matrix

A = (aij)i,j=1...r, aij =2(αi, αj)

(αi, αi)

is called the Cartan matrix of (V,∆).Since any two systems of simple roots are related by an orthogonal

transformation, the matrix A does not depend on any choices up topermutation of indices.

It is also clear that isomorphic root systems have the same Cartanmatrix (up to permutation of indices).

The Cartan matrices determine the root system:

Proposition 22.1. Assume (V1,∆1) and (V2,∆2) are reduced rootsystems with the same Cartan matrix and Span ∆i = Vi. Then theyare isomorphic.

Proof. The assumption means that there exist systems of simple rootsΠ1 ⊂ ∆1 and Π2 ⊂ ∆2 and a linear isomorphism T : V1

∼−→ V2 suchthat TΠ1 = Π2 and such that Tsα = sTαT for all α ∈ Π1. To see thatthe last condition is true, note that

Tsα = Tv − 2(α, v)

(α, α)Tα

andsTαT (v) = Tv − 2(Tα, Tv)

(Tα, Tα)Tα,

and since the Cartan matrices are equal, we can assume that these twoare equal (perhaps up to reordering of some indices).

As the Weyl groups W1,W2 of ∆1,∆2 are generated by simplereflections, we see that

π(ω) = TωT−1

is an isomorphism π : W1 →W2.Recall tnext that ∆1 = ∪ω∈W1

ωΠ1 (and same for ∆2). HenceT∆1 = ∆2 as Tω(α) = π(ω)Tα for all α ∈ Π1.

Finally, for any root β = wα (w ∈W1, α ∈ Π1), we have

Tsβ = Twsαw−1 = π(w)sTαπ(w)−1T = sπ(w)TαT = sTwαT = sTβT.

Instead of the Cartan matrix, we can also represented the rootsystem by its Dynking diagram: it is a (semidirected) graph definedas follows:

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• Its vertices correspond to the set of simple roots α1, . . . , αr.• We connect αi with αj by αijαji edges for i 6= j.• If |aij | < |aji| (i.e. ‖αi‖ > ‖αj‖), then the edge between αi andαj is directed αi → αj .

The Dynkian diagram contains the same information as the Cartanmatrix: first of all aii = 2 for all i. Since the entries are integers, thecondition aijaji = n (n is the number of edges), means that at leastone of aij and aji must be equal to −1. Which one is equal to −1 isdetermined by the direction of the edge between them.

Remark. Note that

aijaji =4(αi, αj)

2

‖αi‖‖αj‖< 4,

by Schwarz’ inequality. Therefore, in the Dynkian diagram, the onlypossible number of edges between two vertices is 0, 1, 2, 3. Note also thatif θij is the angle between αi and αj, then aijaji = 4 cos2 θij. Hencethere are only finitely many possible angles: 90, 60, 45 and 30.

We say that a root system (V,∆) is indecomposable if there areno root systems (V1,∆1), (V2,∆2) with (V,∆) ' (V1 ⊕ V2,∆1 × 0 ∪0 ×∆2).

Lemma 22.2. Show that a root system is indecomposable if and onlyif its Dynkin diagram is connected.

Proof. To come.

Theorem 22.3. All indecomposable, reduced root systems (apart from(R, ∅)), are given by the following diagrams.

Al α1

− α2

− · · · − αl−1

− αl

l ≥ 1 su(l + 1)

Bl α1

− α2

− · · · − αl−1

⇒ αl

l ≥ 2 so(l + 1)

Cl α1

− α2

− · · · − αl−1

⇐ αl

l ≥ 3 sp(2l)

Dl α1

− α2

− · · · −

αl|

αl−2

− αl−1

l ≥ 4 so(2l)

E6 α1

− α2

−

α4

|α3

− α5

− · · · − αl

l = 6, 7, 8

F4 α1

−α2

⇒ α3

− α4

G2 α1

V α2

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To read the detials how it is obtained is maybe not “the most re-warding experience”.

The proof classifies Cartan matrices and implies the following re-sult:

Proposition 22.4. An integer r×r matrix A = (aij)ri,j=1 is a Cartan

matrix of a reduced root system if and only if

1. aii = 2.2. aij < 0, i 6= j.3. There exists a diagonal matrix D with positive entries such that

DA is symmetric and positive definite.

The classification is based on a series of elementary observationabout configurations of vectors in Euclidean speaces. For example, wecan prove the following:

Lemma 22.5. Any Dynking diagram has no loops apart from multipleedges, that is, there are no different vertices α1, . . . , αn (n ≥ 3) withαi and αi+1 connected for i, . . . , n−1, as well as αn and αi connected.

Proof. Consider vi = αi‖αi‖ . Note that 4(vi, vj)

2 = aijaji ∈ 0, 1, 2, 3.Recall also that (vi, vj) ≤ 0 for all i 6= j. Hence either (vi, vj) = 0 or(vi, vj) ≤ − 1

2 for i 6= j. Now consider v =∑i vi. Then

(v, v) = n+∑i 6=j

(vi, vj) ≤ n+ 2

(n−1∑i=1

(vi, vi + 1) + (vn, v1)

)≤ 0

Hence v = 0. But this contradicts the linear independence of theαi.

So the only type of allowed diagrams are trees.

Example 22.6. Let G = SU(n). A maximal torus T is given by theset of diagonal matrices in SU(n). It is isomorphic to Tn−1.

We have the following identifications:

g = su(n) = X ∈ gln(C) | X +X∗ = 0,TrX = 0gC = sln(C) = X ∈ gln(C) | TrX = 0

h = diagonal matrices with trace 0.

Define εi ∈ h∗ by εi(H) = Hii. Then ε1 + . . .+ εn = 0.Then

∆ = αij = εi − αj | i 6= j

with root spaces gαij = CEij , because [H,Eij ] = (Hii −Hjj)Eij .We need an Ad-invariant inner product. One such is given by

(X,Y ) = Tr(XY ). It is Ad-invariant since Ad(g)(X) = gXg−1. For

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X ∈ su(n), we have (X,X) = Tr(X2) = Tr(−X ∗X) < 0 for X 6= 0.Thus (−,−) is negative definite on su(n).

Then hαij = Hαij = Eii −Ejj (as defined above Lemma 20.6). Asa system of simple roots, we can take

α1 = ε1 − ε2, . . . , αi = εi − εi+1, . . . , αn−1 = εn−1 − εn.

This is obviously a system of simple roots. We have

(αi, αk) = (hαi , hαj ) = Tr((Eii−Ei+1,i+1)(Ekk−Ek+1,k+1)) =

2 i = k

−1 i = k ± 1

0 otherwise.

Therefore the Dynking diagram is precisely An−1. F

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23 Semisimple Lie groupsAn ideal a of a Lie algebra g is by definition a subalgebra of g suchthat [X,Y ] ∈ a for all X ∈ g and Y ∈ a.

We say that a Lie algebra is semisimple if it has no non-zeroabelian ideals. A Lie group is semisimple if its Lie algebra is semi-simple.

Lemma 23.1. Suppose G is a connected Lie group and H ⊂ G aconnected Lie subgroup. Then show that h ⊂ g is an ideal if and onlyif H is a normal subgroup.

Proof. Suppose h is an ideal. Then, the condition is that ad(X) ac-tually lands in h. Since ad is the derivative of Ad, this means thatexp(tX)h exp(−tX) ∈ H for all t ∈ R and X ∈ g. But the image ofexp generates G, so this must be true for all of G. Thus H is a normalsubgroup.

The converse is similar.

It follows that a connected Lie group is semisimple if and only ifany normal abelian subgroup of G is discrete.

In particular, if G is a compact connected Lie group, then G issemisimple if and only if any normal abelian subgroup is finite.

Recall that if T ⊂ G is a maximal torus, then N(T ) = T . Thesame argument shows that if a torus T1 is a normal subgroup, thenT1 ⊂ Z(G) (normality implies that we have a homomorphism G →Aut(T1) ' GLk(Z)).

Hence, if A ⊂ G is a normal abelian subgroup, then (A) is a torusnormalized by G, so (A) ⊂ Z(G). Therefore G is semisimple if andonly if Z(G) is finite.

Recall next that the Lie algebra of Z(G) is

z(g) = X ∈ g | [X,Y ] = 0 ∀Y ∈ g.

Thus G is semisimple if and only if z(g) = 0.Now fix a maximal torus T , and consider the corresponding root

decompositiongC = h⊕

⊕α∈∆

hα.

We have z(g)C = z(gC). Assume X = H+∑α∈∆Xα ∈ z(g). Therefore,

for any H ′ ∈ h, we have

0 = [X,H ′] =∑α∈∆

α(H ′)Xα.

Hence Xα = 0 for all α ∈ ∆. In other words, z(gC) ⊂ h.

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Now assume H ∈ z(gC) ⊂ h. We have 0 = [H,X] = α(H)X forX ∈ gα. Hence H ∈ kerα. Thus

z(gC) =⋂α∈∆

kerα.

Thus z(gC) = 0 if and only if⋂α∈∆ kerα = 0 if and only if SpanC ∆ =

h∗.Thus, we have proved the following proposition:

Proposition 23.2. If G is any compact connected Lie group, the fol-lowing are equivalent:

• G is semisimple.

• Z(G) is finite.

• z(g) = 0.

• SpanC ∆ = h∗.

• SpanR ∆ = it∗ ⊂ h∗.

Example 23.3. The center of SU(n) are the scalar matrices of normone. This is isomorphic to Z/n. F

Now let G be a semisimple compact connected Lie group. Fix amaximal torus T ⊂ G and consider the corresponding root system ∆inside ∆ ⊂ it∗.

As a system of simple roots Π ⊂ ∆ form a basis in it∗, the additivesubgroup Q ⊂ it∗ generated by ∆ is a lattice, called the root lattice.Define

P =

β ∈ it∗ | β(Hα) =

2(α, β)

(α, α)∈ Z∀α ∈ ∆

.

We see that P is clearly a subgroup of it∗ with Q ⊂ P . P is alsoa lattice: indeed, if β ∈ P is close to zero, then β(Hα) = 0 (for allα ∈ ∆), so β = 0 since Span ∆ = it∗.

Hence P is a lattice. It is called the weight lattice. In particular,the quotient P/Q is finite abelian.

Example 23.4. Consider the root system A1 corresponding to SU(2).Then it∗ = R. We have that ∆ = −1, 1. Also, P = 1

2Z. HenceP/Q = Z/2. F

Recall that the character latticeX∗(T ) ⊂ it∗ satisfiesQ ⊂ X∗(T ) ⊂P . ((WHY??))

Proposition 23.5. Assume that G is a semisimple compact connectedLie group. Then its univeral cover G is still compact.

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Proof. Let π : G→ G be the covering map.We will need two facts which we will not prove from algebraic topol-

ogy. The first is that kerπ = π1(G). The second is if G is a compactmanifold, then π1(G) is finitely generated as a group (this follows fromthe van Kampen theorem).

As kerπ is a normal discrete subgroup of G, we have kerπ ⊂ Z(G).Hence kerπ ' Zn ⊗ Γ where Γ is a finite abelian group. We want toshow that n = 0. This will imply the statement since then the coveringis finite. So assume that n > 0.

Then we can find a subgroup H ⊂ kerπ such that kerπ/H is finite,but arbitrarily large. Then consider G1 = G/H. We have a homomor-phism π1 : G1 → G with kerπ1 = kerπ/H, hence G1 is compact, butkerπ1 can be chosen arbitrarily large. This will lead to a contradiction.

Fix a maximal torus T ⊂ G. We identify g1 with g using (π1)∗.Consider the maximal torus π1 = expG1

t ⊂ G1.Consider the root and weight lattices Q,P ⊂ it∗ = it∗1. We have

Q ⊂ X∗(T ) ⊂ X∗(T1) ⊂ P . Hence

|X∗(T1)/X∗(T )| ≤ |P/Q|.

One can show that kerπ1 = X∗(T1)/X∗(T ). Hence

| kerπ1| = |X∗(T1)/X∗(T )| ≤ |P/Q|.

Since the left-hand-side can be arbitrarily large, this is a contradiction,since P/Q is finite.

Let Ad(G) be the Lie group G/Z(G). Then Ad(G) is semisimple,because ........

Proposition 23.6. Assume G is a compact connected Lie group. Theng = z(g)⊕ [g, g].

Proposition 23.7. Any compact connected Lie group G can be writtenas (Tn×H)/Γ, where H is a compact simply connected semisimple Liegroup and Γ is a finite subgroup of Tn × Z(H).

Proof. The group G/Z(G) is semisimple with Lie algebra g/z(g) '[g, g].

Consider the universal cover H of G/Z(G). Then H is compactand semisimple and simply-connected. The identification of the Liealgebra of H with [g, g] ⊂ g defines a homomorphism H

π−→ G. Thenwe get a homomorphism π′ : Z(G)×H → G given by (g, h) 7→ gπ(h).At the Lie algebra level, this is an isomorphism.

Hence π′ is surjective and Γ = kerπ′ is finite.

92

24 Classification of simply connectedsemisimple Lie groupsLet G be a compact connected Lie group. Fix a maximal torus T ⊂ G.Consider a finite-dimensional representation π : G → GL(V ). We candecompose π

∣∣Tinto isotypic components. Thus, for λ ∈ X∗(T ), define

V (λ) = v ∈ V | Hv = λ(H)v ∀H ∈ h.

The elements of V (λ) are called vectors of weight λ. Put

P (V ) = λ ∈ h∗ |V (λ) 6= 0 ⊂ X∗(T ).

ThenV =

⊕λ∈P (V )

V (λ).

If X ∈ gα, then XV (λ) = V (λ+ α). Indeed, for H ∈ h, v ∈ V (λ),we have

HXv = (HX −XH)v +XHv

= [H,X]v +XHv

= α(H)Xv + λ(H)Xv

= (α(H) + λ(H))Xv.

Proposition 24.1. Let λ ∈ P (V ) and α ∈ ∆. Then

1. The set λ+ kα ∈ P (V ) | k ∈ Z have the form

λ− qα, λ− (q − 1)α, . . . , λ+ pα

for some p, q ≥ 0 such that p− q = λ(Hα).2. Ep+qα V (λ− qα) = V (λ+ pα).

Proof. Consider the copy of sl2(C) ⊂ gC spanned by g−α, gα and Hα.Consider V as a sl2(C)-module. From the representation theory ofsl2(C), we know that the set

λ(Hα) + 2k | λ+ kα ∈ P (V )

has the form−n,−n+ 2, . . . , n− 2, n

for some integer n = 2s. This means that

λ+ kα | P (V ) = λ− qα, . . . , λ+ pα.

Hence −n = λ(Hα)− 2q and n = λ(Hα) + 2p. Adding these equationsgive p− q = λ(Hα).

The second property follows since E raises the indices and thatE2s−sξ−s = ξs, up to scalar.

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Corollary 24.2. If λ, λ+α ∈ P (V ) for some α ∈ ∆, then EαV (λ) 6=0.

In particular, applying this to the adjoint representation, we con-clude that if α, β and α+ β ∈ ∆, then [gα, gβ ] = gα+β : we know fromearlier than the left-hand-side is contained in the right-hand-side. Bythe above corollary, it is nonzero, and both sides are one-dimensional.

Proposition 24.3. Fix a system of simple roots Π ⊂ ∆. Then everyα ∈ ∆+ can be written as αi1 + . . . + aik such that αij ∈ Π andαi1 + . . .+ αil ∈ ∆ for all l = 1, . . . , k.

Proof. FFF

Corollary 24.4. The space gC is generated by h and Eα, Fα for α ∈ Π.

Proof. If λ = αi1 + . . . + αik ∈ ∆+ as in the proposition, then gλ =[[[Eα1

, Eα2], Eα3

], . . . , Eαk ]⊗ C.

Theorem 24.5. Two compact simply connected semisimple Lie groupsare isomorphic if and only if their root systems are isomorphic.

Proof. Fix a maximal torus T and a system of simple roots Π =α1, . . . , αr. Write Hi, Ei, Fi for Hαi , Eαi , Fαi . Let A denote theCartan matrix A = (aij)

ni,j=1, with aij = 2(αi, αj)/(αi, αi).

Let a be the universal Lie algebra with generators Hi, EiFi fori = 1, . . . , r subject to the relations

[Hi, Hi] = 0 [Ei, Fi] = Hi

[Hi, Ej ] = aijEj [Ei, Fj ] = 0(i 6= j)

[Hi, Fj ] = aijFj

Then we get a Lie algebra homomorphism π : a → gC. We needto describe kerπ. Denote by h = SpanHii=1,...,r, and by a+ ⊂ a

the subalgebra generated by Ei (i = 1, . . . , r), and by a− ⊂ a thesubalgebra generated by Fi (i = 1, . . . , r).

Then it is not difficult to see that a = a−⊕ h⊕a+ as a vector space.Define αi ∈ h∗ by αi(Hj) = aji.Then we can define root spaces

aλ = X ∈ a | [H, X] = λ(H)X ∀ H ∈ h.

Then a+ and a− decomposes into positive and negative root spaces,respectively. That is:

a+ =⊕

λ=n1α1+...nrαr, ni∈Z+

aλ and a− =⊕

λ=n1α1+...nrαr, ni∈Z−

aλ

94

Now, if p ⊂ a is an ideal8,then p =⊕

λ(aλ ∩ p). It follows that thesum of all ideals p in a such that p ∩ h = 0 still is an ideal m with thesame property, that is, m ∩ h = 0.

The claim is now that ker(π : a→ gC) = m. Thus gC = a/m.As π : a→ gC gives an isomorphism h ' h, we have that kerπ∩ h =

0. Hence kerπ ⊂ m.Consider the ideal p = m/ kerπ in gC. Then p ∩ h = 0. Then

p =⊕α∈∆

(p ∩ gα).

If p 6= 0, then p ∩ gα 6= 0 for some α ∈ ∆, so Eα ∈ p. But thenHα = [Eα, Fα] ∈ p ∩ h, which is a contradiction. Hence kerπ = m.

Therefore gC = a/m, so gC is described entirely in terms of theCartan matrix.

Recall also from earlier that we can recover g from gC as the sub-space

X ∈ gC | X? = −X,

where X? was the involution introduced earlier. The involution iscompletely determined by what it does on generators, so we can recoverg.

Finally, we conclude that if G1, G2 are two simply-connected com-pact Lie groups with isomorphic root systems, then their Cartan ma-trices coincide, hence g1 ' g2, and therefore G1 ' G2.

In addition to the obvious relations in g, we have the so-calledSerre relations:

Proposition 24.6 (Fact). The relations

(adEj)1−aji(Ei) = 0 (adFj)

1−aji(Fi) = 0 i 6= j

together with relations used in the proof are relations in gC.

As mentioned earlier, conversely, any reduced root system (V,∆)with Span ∆ = V arises from a (simply-connected) compact semisimpleLie group. This can be shown in different ways: one way is as in theproof above. Define gC = a/m as above and integrate. The problemin this approach is to show that gC is finite-dimensional (and gives riseto a finite-dimensional Lie group).

The other approach is to use the classification of reduced root sys-tems and construct each Lie group corresponding to each root system.Those missing are the 5 exceptional ones.

8Keep in mind that being an ideal is the same as being an ad h-invariant subspace

95

25 Representation theory of compact con-nected Lie groupsLet G be a compact connected Lie group. Fix a maximal torus T ⊂ G.Fix a system of simple roots Π = α1, . . . , αr ⊂ ∆. Then for anyfinite-dimensional representation Vπ of G, we have a weight decompo-sition

V =⊕

λ∈P (V )

V (λ).

where V (λ) = v ∈ V | Hv = λ(H)v ∀ H ∈ h 6= 0, andP (V ) ⊂ X∗(T ) ⊂ it∗ ⊂ h∗.

We saw in the last lecture that EαV (λ) ⊂ V (λ+α) and FαV (λ) ⊂V (λ− α).

We call a nonzero vector v ∈ V (λ) a highest weight vector (ofweight λ) if Eαv = 0 for all α ∈ ∆+. Equivalently, if Eiv = 0 for alli = 1, . . . , r.

If in addition, v generated V as a gC-module, then V is a called ahighest weight module.

Clearly, if V is an irreducible representation, then it is a heighestweight module: the highest weight is the largest element in P (V ) withrespect to the partial order

µ ≥ η ⇔ η = µ− αi1 − . . .− αik .

Thus any irreducible finite-dimensional representation of G has auniquely assigned weight λ ∈ X∗(T ).

Recalling the representation theory of sl2(C), we also know thatλ(Hi) ≥ 0 for all i = 1, . . . , r.

A weight λ ∈ X∗(T ) such that λ(Hi) ≥ 0 for all i = 1, . . . , r iscalled dominant.

Theorem 25.1. For any compact connected Lie group with maximaltorus T and a fixed system of simple roots, we have:

1. For every dominant weight λ ∈ X∗(T ), there exists an irreduciblefinite-dimensional representation πλ of G of highest weight λ.

2. Any finite-dimensional irreducible representation of G is equiva-lent to πλ for a uniquely defined λ.

We will prove this only for the simplyconnected semisimple case.Simultaniously, we will prove that X∗(T ) = P .

Proof. Here is a sketch of the proof.We know that SpanC ∆ = h∗ and Q ⊂ X∗(T ) ⊂ P .Recall that

P =

β ∈ it∗ | β(Hα) =

2(α, β)

(α, α)∈ Z∀α ∈ ∆

.

96

and that Q = SpanZ ∆. The set of dominant weights in P , that is, theset

λ ∈ P | (λ, αi) ≥ 0∀i

is denoted by P+. The elements of P+ are called dominant integralweights. Note that P+ = P ∩ C.

The goal is to show that for any λ ∈ P+, there exist a uniquefinite-dimensional irreducible highest weight gC-module of weight λ.

This will prove the theorem in the semisimple simplyconnectedcase, as well that X∗(T ) = P .

Denote by Ug the universal enveloping algebra of g. The Poincaré-Birkhoff-Witt-theorem says that if X1, . . . , Xn is a basis of g, thenXa1

1 · · ·Xann with a1, . . . , an ≥ 0 is a basis of Ug.

This implies that the map g → Ug is injective, so we can considerg as a subspace of Ug. If a, b ⊂ g are Lie algebras and g = a ⊕ bas a vector space, then the homomorphisms Ua → Ub and Ub → Uginduce a linear isomorphism Ua⊗U b→ Ug.

Returning to semisimple simplyconnected Lie groups, we have

gC = h⊕⊕α∈∆

gα.

Consider the following two Lie subalgebras in gC:

a− =⊕α∈∆−

gα b = h⊕⊕α∈∆+

gα.

The module b is called a Borel subalgebra. Then U(gC) = Ua− ⊗CUb.

For any λ ∈ h∗, we have a 1-dimensional representation ρλ: ρλ :Ub → C defined by ρλ(H) = λ(H), and ρλ(gα) = 0. Denote thisrepresentation by Cλ.

Then we define a UgC-module by induction:

Lλ = UgC ⊗Ub Cλ.

Note that, as vector spaces, Lλ ' Ua−. Put vλ = 1 ⊗ 1 ∈ Lλ.Then vλ generates Lλ and Hvλ = λ(H)vλ and Eαvλ = 0. Thus Lλ isa highest weight module with highest weight vector λ. It is called aVerma module.

It has the following universal property: if V is a highest weightmodule with highest weight vecotr w ∈ V (λ), then there exists a uniquemorphism Lλ → V such that vλ 7→ w. Indeed, Cλ → V (1 7→ w) is amorphism of b-modules. Then we define Lλ → V to be the composition

Lλ = U(gC)⊗Ub Cλ → U(gC)⊗Ub V → V.

97

We have a weight decomposition

Lλ =⊕µ≤λ

Lλ(µ).

We can see that dimLλ(µ) is equal to the number of ways of writingλ − µ as

∑mj=1 ajβj for aj ∈ Z+. This number is called Kostants

partition function of λ−µ. In particular, dimLλ(µ) is always finite.For any submodule M ⊂ Lλ we have

M =⊕µ≤λ

(M ∩ Lλ(µ)).

It follows that the sum of all submodules Mλ such that vλ 6∈ Mstill has the same property.

Now define Vλ = Lλ/Mλ.Denote by ξλ the image of vλ in Vλ.

Lemma 25.2. For any λ ∈ h∗, Vλ is a simple module. Furthermoe,any simple highest weight module is isomorphic to Vλ for a uniquelydefined λ ∈ h∗.

Proof inside proof: If there is a proper nonzero submodule M of Vλ,then it lifts to proper inclusions Mλ ⊂ M ⊂ Lλ. We cannot havevλ ∈M , for then M = Lλ (since vλ generates Lλ). Hence vλ 6∈M , soMλ = M by maximality of Mλ. Thus Vλ is simple.

Now suppose that V is a simple highest weight module with highestweight vector v ∈ V (λ). Then we have a morphism π : Lλ → V givenby vλ 7→ w. But λ is simple, so this must be an isomorphism.

The highest weight is uniquely determined, as all other weights ofV are lower than λ.

As any finite-dimensional simple module is a highest weight mod-ule, to finish the classification of such modules (thus finite-dimensionalrepresentations of G), it remains to understand when Vλ is finite-dimensional.

Proposition 25.3. The module Vλ is finite-dimensional if and only ifλ ∈ P+.

Proof. This is long and tricky. The hardest part.

Once the above proposition is proved, we see that the Vλ for λ ∈ P+

are all irreducible representations of g.

Proposition 25.4. Any finite-dimensional highest weight module issimple, hence isomorphic to Vλ for some λ ∈ P+.

98

26 Classical approach to representationtheory of SU(n)

Recall that the standard chouce of a maximal torus is

T =

z1

. . .

zn

| zi ∈ T,∏

zi = 1

.

The Lie algebra is

su(n) = X ∈ gln(C) | TrX = 0, X +X∗ = 0.

The complexification of su(n) is identified with sln(C), which consistof complex matrices with trace zero.

The Cartan subalgebra is

h =

H =

H11

. . .

Hnn

|∑i

Hii = 0

.

Define Li ∈ h∗ by Li(H) = Hii. Then L1 + . . .+ Ln = 0, and

∆ = Li − Lj | i 6= j.

As a system of simple roots we can take Π to be αi = Li − Li+1 fori = 1, . . . , n − 1. The root spaces are gLi−Lj = CEij , where Eij isthe matrix (δij). As an Ad-invariant inner product on h, we may take(H,H ′) = Tr(HH ′).

Recall that the isomorphism h∗ ' h, α 7→ hα is defined by (H,hα) =α(H) for all H. For α = Li we get Hii = Tr(HhLi) for i = 1, . . . , n.By inspection we see that then we must have hLi = Eii − 1

nI. Thenthe form on h∗ is given by

(Li, Lj) = (hLi , hLj ) = Tr((Eii −1

nI)(Ejj −

1

nI)) = δij −

1

n.

Hence (αi, αi) = (Li − Li+1, Li − Li+1) = 1− 1n + 2

n + 1− 1n = 2.

ThusHαi = α∨i =

2hα(αi, αi)

= Eii − Ei+1,i+1,

which we can identify with αi.The fundamental weights ωi are defined by (ωi, α

∨j ) = δij . From

this we see that

99

ωi = L1 + . . .+ Li, i = 1, . . . , n− 1.

It follows that P = Zω1 + . . . + Zωn−1 = ZL1 + . . . + ZLn−1 andthat P+ = Z+ω1 + . . .+ Z+ωn−1.

Put V = Cn. Consider the obvious representation π of G = SU(n)on V . It is obviously irreducible.

The vector e1 = (1, . . . , 0) is a highest weight vector, Eije1 = 0 fori < j of weight ω1 = L1, as He1 = H11e1. Thus π = πω1

= πL1.

The representations corresponding to the fundamental weights arecalled fundamental. Since G is a compact subgroup of SL(V ), weknow that every irreducible representation of G is a subrepresentationof π⊗kω1

for some k ≥ 0. How do we explicitly realize πa1L1+...anLN

(a1 ∈ Z+, a1 ≥ a2 ≥ . . . ≥ an−1) as a subrepresentation of π⊗kω1?

Fix k ≥ 2. Consider the representation σk of Sk on V ⊗k defined by

σk(τ)(v1 ⊗ . . .⊗ vk) = vτ−1(1) ⊗ . . .⊗ vτ−1(k).

Theorem 26.1 (Schur-Weyl duality). The subalgebras π⊗k(SU(n))and σk(Sk) of End(V ⊗k) are commutants of each other.

[[insert proof]]Let ρi : Sk → GL(Vi) be the different irreducible representations

contained in σk. We then have a decomposition into isotypic compo-nents:

V ⊗k =

m⊕i=1

Vi ⊗Wi,

with σk(τ) =⊕m

i=1 ρi(τ)⊗ 1.Then σk(C[Sk]) =

⊕1⊗End(Wi). It follows that we get represen-

tations ρi : SU(n)→ GL(Wi) such that .........Therefore we get a 1-1 correspondence between the irreducible rep-

resentations of Sk contained in σk and the irreducible representationsof SU(n) contained in π⊗kω1 .

Schur functor

Remark. To complete the classification of irreducible representationsof SU(n) without relying on Verma modules and such, it would remainto show that two irreducible representations with highest weight areequivalent, that is, Sλ(V ) = Sλ

′(V ) if λi = λ′i + a for some a ∈ Z.

100

A Exercises

A.1 Lecture 1Exercise 1. Suppose V1⊕ . . .⊕Vr ∼W1⊕· · ·⊕Ws as representationsand that the Vi,Wi are irreducible. Show that n = m and there existsρ ∈ Sr such that Vi ∼Wρ(i) for all i. ♠

Solution 1. First note a lemma: If V ⊕W ∼ V ′ ⊕W ′ and V ∼ V ′,then W ∼W ′. This follows from the five-lemma.

Now consider the composition V1 → V1⊕· · ·⊕Vr →W1⊕· · ·⊕Ws →Wi. This has to be non-zero for at least one i. Hence V1 ∼ Wi forsome i. By rearring, we are in the situation of the lemma above. Hence,inductively, if r ≤ s, we find 0 ∼Wr+1⊕· · ·⊕Ws, which is impossible.Similarly for s ≤ r. Hence r = s and we conclude. ♥

Exercise 2. Let G denote the Pontryagin dual of G. Then, for anyfinite abelian group G, we have G ' G. ♠

Solution 2. We first do this when G = Z/n for some n ≥ 0. Letϕ : Z/n → C∗ be a character. Then φ(1) = z for some z, but φ(n) =φ(0) = zn = 1, so z must be a n’th root of unity.

Realize Z/n as the nth roots of unity. Then we can define a ho-momorphism Z/n → Z/n by ϕ 7→ ϕ(1). Similarly, we can define aninverse map by sending e2πi/n to the character m 7→ e2πim/n. Theseare inverses.

Now, every finite abelian group is a product of these groups. Soit remains to show (by induction) that if G,H are two finite abeliangroups, then G×H ' G × H. The inclusion maps G → G × H andH → G×H induces a map G× H → G×H by (ϕ1, ϕ2) 7→ ((g, h) 7→ϕ1(g)ϕ2(h)).

It is easy to see that this map is injective. To see that it is surjective,let ϕ : G × H → C be a character. Write ϕ as gh 7→ ϕ(gh). Thenwe can define characters on G,H by g 7→ ϕ(g · 1) which maps to ϕ.Ok. ♥

A.2 Exercises 3Exercise 3. Show that the S defined in the proof of Theorem (??)lies in Θ(G)′′. ♠

Solution 3. This is, I think, definition-hunting. DETAILS COMELATER ♥

Exercise 4. 1. Check that indeed

〈v, w〉 :=1

|G|∑g∈G〈π(g)v, π(g)w〉′

101

is an invariant Hermitian scalar product.

2. Show that if π is irreducible, then an invariant scalar product isunique up to a factor.

♠

Solution 4. i). Invariant means that 〈π(h)v, π(h)w〉 = 〈v, w〉. This isclear from the definitions, since if g ∈ G ranges over all of G, then sodoes gh.

The only (slightly) nontrivial thing to check is that 〈, 〉 is positivedefinite. But this is so.

ii) If 〈, 〉 is invariant, then it is an element of HomG(V ⊗V,C). Thisis canonically isomorphism to HomG(V⊗, V ∗). If we can show that Vand V ∗ are isomorphic as representations, then we are done by Schur’slemma.

But we are given an inner product 〈, 〉. We can define a map V →V ∗ by sending v ∈ V to the function w 7→ 〈v, w〉. This is clearly alinear isomorphism, and it is also a map of representations. Recallthat the action of G on V ∗ is defined by π∗(g)ϕ(v) = ϕ(g−1v). Thenin the diagram

V //

·g

V ∗

·g

V // V ∗

we want 〈gv, w〉 to be equal to 〈v, g−1w〉. But this is true since 〈, 〉 isinvariant:

〈gv, w〉 = 〈gvgg−1w〉 = 〈v, g−1w〉.

Hence V and V ∗ are isomorphic as representations and thenHomG(V, V ∗) (space of bilinear forms) is one-dimensional. ♥

Exercise 5. Let (V, π) be an irreducible representation and (W, θ) afinite-dimensional representation. LetW (π) be the isotypic componentcorresponding to π.

1. Show that the operator

P =dimπ

|G|∑g∈G

χπ(g−1)θ(g)

on W is the projection onto W (π) along∑π′ 6'πW (π′). In par-

ticular:W =

⊕[π]∈G

W (π).

2. Show that θ∣∣W (π)

∼ πnπ where nπ = dimMor(π, θ).

102

♠

Solution 5. dddd ♥

A.3 Lecture 4Exercise 6. 1. πcc = π.

2. (pi⊗ θ)c = πc ⊗ πc.3. π is irreducible if and only if πc is.

♠

A.4 Exercises 5Exercise 7 (Excercise 1). Show that the ring Z is integrally closed. ♠

Solution 6. Let x = p/q ∈ Q and suppose that we have an equationof integral dependence:

xn + b1xn−1 + . . .+ bn = 0

with bi ∈ Z. We can suppose that p and q have no common factor.Multiply by qn on both sides to get

pn + b1qpn−1 + . . .+ bnq

n = 0.

Looking modulo q we see that q | pn, which to avoid contradiction,must imply that q = ±1, hence x ∈ Z. ♥

Exercise 8 (Exercise 2). Suppose R is a unital commutative ring andS ⊂ R a subring with 1 ∈ S. Show that the set of elements of Rintegral over S form a subring of R. ♠

Solution 7. Note that x is integral over S if and only if S[x] is afinitely generated S-module. Hence if x and y are integral over S, itfollows that S[x] ⊗S S[y] = S[x, y] is a finitely generated S-module.Hence x± y and xy are integral over S ♥

A.5 Exercises 11

A.6 Lecture 13Exercise 9. Let G = GLn(C). Show that exp : g → G is surjective.Show also that it is not open. ♠

Solution 8. Note that for matrix groups, the exponential is given by

exp(A) = I +A+1

2A2 +

1

6+A3 + . . .+

1

n!An + . . .

103

Also note thatexp(P−1AP ) = P−1 exp(A)P

for invertible matrices P ∈ GLn(C) ⊂ g = End(Cn). Thus, if A is diag-onalizable with eigenvalues λi, the matrix P−1diag(log λ1, . . . , log λn)P(with P diagonalizing A) is mapped to A. Here log is of course the com-plex logarithm (which is multi-valued, so choose one value for each).

Now if A is not diagonalizable, the matrix A + εI is invertibleand diagonalizable for generic ε. Thus we can find a sequence Anwith all An diagonalizable and An → A with corresponding Bn withexp(Bn) = An. I claim that the Bn can be chosen such that Bnis a convergent sequence. This is because the eigenvalues of An arecontinous with respect to ε, and as log : C∗ → C is continous, wecan choose the logs of close eigenvalues to be close. Thus we get aconvergent sequence Bn → B, and by continuity of exp, exp(B) = A.

Now we explain why exp is not open. [[[HOW????]] ♥

A.7 Exercises 13Exercise 10 (Exercise 1). Let G be a group that is also a compactspace (that is, a compact Hausdorff topological space) such that themultiplication map µ : G × G → G is continous. Show that G is acompact group, that is, that the inverse map ι : G → G is continousas well. ♠Solution 9. Let U ⊂ G. Then

ι−1(U) = g ∈ G | ι(g) ∈ U= g ∈ G | g−1 ∈ U= g ∈ G | e ∈ gU

[[[[[[[[[ something about bijection between compact spaces is a home-omorphism ]]]]]]]]] ♥Exercise 11. 1. Find a Haar measure on GL2(R) and check that

this group is unimodular9.2. Find left- and right-invariant Haar measures on the ax+ b group

over R, and check that this group is not unimodular.♠

Solution 10. For the first part, we start with the usual measure onR4, and see what happens when we translate by an element of GL2(R).

We assume the Haar measure has the following form, for some yetunknown function f .

µ(S) =

∫S

f(A)|da11da12da21daa22|.

9Recall that this means that the left and right invariant Haar measures coincide

104

Now consider the same formula on the set gS. Then the change ofvariable formula tells us that

µ(gS) =

∫S

f(gS)|detD(g)||da11da12da21daa22|.

where detD(g) is the Jacobian matrix of the map A 7→ gA. A compu-tation shows that this map is row-equivalent to a block diagonal matrixwith two copies of g. Thus detD(g) = (det g)2. Hence a left-invariantHaar measure on GL2(R) is given by

µ(S) =

∫S

1

|detA|2dA.

The same computation with g 7→ Ag gives the same result. HenceGL2(R) is unimodular.

ii). For the second part, note that the ax+ b-group is topologicallyR × R∗, so that we can use the measure dxdy here. Then a similarcomputation gives that a left-invariant Haar measure is given by

µ(S) =

∫S

1

a2da db

and a right-invariant Haar measure is given by

µ(S) =

∫S

1

ada db.

♥

A.8 Exercises 16Exercise 12 (Exc 2). Let G be a topological group and Γ ⊂ G adiscrete normal subgroup. Show that Γ ⊂ Z(G). ♠Solution 11. Since Γ ⊂ G is normal, we have that h−1gh ∈ Γ for allg ∈ G. So we get a continous map c : G → Γ given by conjugation.But a continous map from a connected topological space to a discretespace must be constant (let g1, g2 ⊂ c(G), which is open (in thediscrete topology), hence c−1(g1, g2) is open, which is only possibleif the inverse image is ∅ or all of G).

Since c is constant it must be constantly equal to e, since c(e) =e. ♥

A.9 Exercises 18Exercise 13 (Exc 2). Let π : G → GL(V ) and η : G → GL(W ) betwo finite-dimensional representations of a Lie group G. Show that

(π ⊗ η)∗(X) = π∗(X)⊗ 1 + 1⊗ η∗(X)

for all X ∈ g. ♠

105

Solution 12. We have that GL(V ⊗W ) is an open subset of End(V ⊗W ), so that we may use the additive structure when computing thederivative.

Thus, let γ : R→ GL(V ⊗W ) be a continous curve with γ(0) = eand γ′(0) = X. Then

π∗(π ⊗ η)(X) =∂

∂t(π(γ(t))⊗ η(γ(t)))

∣∣t=0

= limt→0

1

t(π(γ(t))⊗ η(γ(t))− π(γ(t))⊗ 1 + π(γ(t))⊗ 1− 1⊗ 1)

= 1⊗X +X ⊗ 1

by the usual proof of the product rule. ♥

A.10 Exercises 19Exercise 14 (Exercise 3). Let G be a closed subgroup of U(n). Verifythat the formula

〈X,Y 〉 = −Tr(XY )

defines an Ad-invariant scalar product on g ⊂ u(n). ♠

Solution 13. We first verify that the formula defines a scalar product.It is clearly symmetric and bilinear. We only need to check that it ispositive definite. Note that every X ∈ g satisfies X = −X∗. Hence〈X,X〉 = −Tr(XX) = −Tr(−X∗X) = Tr(XX∗). An easy computa-tion shows that the ith diagonal entry of XX∗ is the sum of the moduliof the elements in the ith row of X. Hence tr(XX∗) is the sum of thelengths of all the entries in X (with certain repetitions). But this iszero if and only if X = 0. Hence 〈X,Y 〉 is positive definite.

Recall the adjoint action of G on g, given by g ·X = gXg−1. BeingAd-invariant means that 〈g · X, g · Y 〉 =〉X,Y 〈 for all X,Y ∈ g. Butthe action on the Lie algebra is given by conjugating, and we have

〈g·X, ·Y 〉 = 〈gXg−1, gY g−1〉 = −Tr(gXg−1gY g−1) = −Tr(gXY g−1) = −tr(XY ).

And we are done. ♥

106

B Some worked examples

B.1 The representation ring of S3

We want to explicitly compute the representation ring of S3.First we find all irreducible representations. The first one is the

trivial representation ε : S3 → C. We also have the sign represen-tation given by g 7→ sgn(g) ∈ C. Both of these are one-dimensionalrepresentations.

S3 have a natural action on R3 given by permuting the basis vec-tors. But it acts trivially on the subspace spanned by e1 + e2 + e3,hence R3 decomposes as ε ⊕ V , where V is some 2-dimensional rep-resentation. It is irreducible: if not, any v ∈ V would be sent to ascalar multiple of itself, but this is not the case. This representation iscalled the standard representation of S3. We have also now foundall representations of S3, since 12 + 12 + 22 = 6.

Thus the representation ring R(S3) is Z[A,S] modulo some rela-tions to be found. Here A is the alternating representation and S isthe standard representation. The trivial representation is 1. It is easyto see that A ⊗ A ∼ ε, so A2 = 1 ∈ R(S3). The representation S ⊗ Sis 4-dimensional. To compute how it decomposes, we use charactertheory.

Recall that characters are class functions on G, that is, they onlydepend on the conjugacy classes of G. So we write a character table:

e τ σ

ε 1 1 1

A 1 1 -1

S 2 -1 0

To compute the character of the standard representation, we firstnote that it is, as a vector space given by R3/(1, 1, 1). A basis isthen given by the images of e1, e2. Let τ be the transposition (123),sending ei to ei+1. Let σ be reflection fixing e1 and exchanging e2 ande3. In this basis that means e2 is sent to −e1 − e2. Writing up thecorresponding matrices lets us find the value of the character.

Now one can compute by hand (or use a result on characters), thatthe character of S ⊗ S is given by χS · χS , so that its entry in thecharacter table is (4, 1, 0). If S ⊗ S = V1 ⊕ V2, then χS = χV1

+ χV2.

Using also that the characters are linearly independent, we see thatthe only option is S ⊗ S ∼ ε⊕A⊕ S.

Hence S2 = 1 +A+S in the representation ring. Similarly, we findthat AS = S in the representation ring. All in all

R(S3) = Z[A,S]/(A2 − 1, S2 − 1−A− S,AS − S).

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B.2 Explicit Specht modulesAgain, we work with G = S3. Consider the following Young-diagram:

T =

1 23

We want to use the theorem from Lecture 5 (?) to find the standardrepresentation of S3. The elements of S3 are generated by ρ = (123)and s = (23) with the relations ρ3 = e and sρs = ρ2.

First off, the elements permuting the rows is the subgroup consist-ing of e and (12) = sρ2. The elements permuting the columns is thesubgroup generated by (13) = sρ. Then

aT =1

2(e+ (12))

andbT =

1

2(e+ (13)) .

ThuscT =

1

4(e− (13) + (12)− (132)) .

This gives us as in the lecture a map

C[S3]→ V ⊆ C[S3]

given by g 7→ gcT , whose image is supposed to be an irreducible repre-sentation of S3. Let’s find this. One computes the action of cT on thebasis elements of C[S3] by direct computation:

cT =1

4

(e− sρ+ sρ2 − ρ2

)ρcT =

1

4(ρ− s+ sρ− e)

ρ2cT =1

4

(ρ2 − sρ2 + s− ρ

)scT =

1

4

(s− ρ+ ρ2 − sρ2

)sρcT =

1

4(sρ− e+ ρ− s)

sρ2cT =1

4

(sρ2 − ρ2 + e− sρ

)

We claim that cT and ρcT span the image. For we have ρ2cT = −cT −ρcT . And scT = ρ2cT . And sρcT = ρcT . And sρ2cT = cT .

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Thus we have some 2-dimensional representation of S3 with basiscT , ρcT . We compute the matrices of S3 with respect to this basis:

e =

1 0

0 1

ρ =

0 −1

1 −1

ρ2 =

−1 1

−1 0

s =

−1 0

−1 1

sρ =

0 1

1 0

sρ2 =

1 −1

0 −1

.

As expected, the character of this representation is exactly the char-acter of the standard representation from the previous example (as isseen by computing traces). This is no coincidence.

References[1] M. F. Atiyah and I. G. Macdonald. Introduction to commutative

algebra. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1969.

[2] Allen Hatcher. Algebraic topology. Cambridge University Press,Cambridge, 2002.

[3] Jean-Pierre Serre. Linear representations of finite groups. Springer-Verlag, New York-Heidelberg, 1977. Translated from the secondFrench edition by Leonard L. Scott, Graduate Texts in Mathemat-ics, Vol. 42.

[4] Michael Spivak. A comprehensive introduction to differential ge-ometry. Vol. I. Publish or Perish, Inc., Wilmington, Del., secondedition, 1979.

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